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Since we are asked "find the potential of the shell", we must assume that it is not grounded (so it is not kept at $V=0$) and that means the total charge on the shell remains $0$. If not, the net charge on the shell would be an extra free parameter, $Q_s$, which would have to be specified before we can answer the questions.

On the outside we don't see what happens inside, but only that the total charge contained is $+Q$. We can now:

1. Distribute the internal charge $+Q$ equally over the surface, without the external $Q$ present. That gives $V({\bf x})=Q/(4\pi\varepsilon_0|{\bf x}|)$.
2. Subsequently put the external $Q$ at its position ${\bf x}_{\rm ext}=(3R,0,0)$ and give it the correct image charge $Q_{\rm ext\_im}=-2Q/3$ inside the sphere at location ${\bf x}_{\rm ext\_im}=(4R/3,0,0)$
3. Compensate for the image charge because the sphere is not grounded and has to remain chargeless, by adding $-Q_{\rm ext\_im}$ uniformly distributed at the surface.

So the external potential is now known: $$ V_\text{ext}({\bf x}) = \frac{5Q/3}{4\pi\varepsilon_0|{\bf x}|} +\frac{-2Q/3}{4\pi\varepsilon_0|{\bf x}-{\bf x}_{\rm ext\_im}|} +\frac{Q}{4\pi\varepsilon_0|{\bf x}-{\bf x}_{\rm ext}|} $$ which gives the potential at (any point of) the surface of the shell as: $$ V_\text{shell} = \frac{5Q}{24\,\pi\varepsilon_0\,R} \tag{1} $$

On the inside we don't see what happens at the outside, only that the shell is at potential $V_\text{shell}$. We also have the charge $Q$ at position ${\bf x}_{\rm int}=(-R/2,0,0)$ and we give it an image charge $-2Q$ at position ${\bf x}_{\rm int\_im}=(-2R,0,0)$ so the potential in the cavity is: $$ V_\text{int}({\bf x}) = V_\text{shell}+\frac{Q}{4\pi\varepsilon_0|{\bf x}-{\bf x}_{\rm int}|} +\frac{-2Q}{4\pi\varepsilon_0|{\bf x}-{\bf x}_{\rm int\_im}|} $$ which gives the potential in the center as: $$ V_\text{cen} = \frac{11Q}{24\,\pi\varepsilon_0\,R} \tag{2} $$ NB: In the above we assume that the reader is accustomed with using [image charges] for a sphere, which always scale with the relative distance to the spherical image surface, and of course have opposite sign.

PS: The question's title was changed in an edit by OP, it previously didn't contain "without images", that's why this answer was valid. (And of course it gives the right numbers.)

Since we are asked "find the potential of the shell", we must assume that it is not grounded (so it is not kept at $V=0$) and that means the total charge on the shell remains $0$. If not, the net charge on the shell would be an extra free parameter, $Q_s$, which would have to be specified before we can answer the questions.

On the outside we don't see what happens inside, but only that the total charge contained is $+Q$. We can now:

1. Distribute the internal charge $+Q$ equally over the surface, without the external $Q$ present. That gives $V({\bf x})=Q/(4\pi\varepsilon_0|{\bf x}|)$.
2. Subsequently put the external $Q$ at its position ${\bf x}_{\rm ext}=(3R,0,0)$ and give it the correct image charge $Q_{\rm ext\_im}=-2Q/3$ inside the sphere at location ${\bf x}_{\rm ext\_im}=(4R/3,0,0)$
3. Compensate for the image charge because the sphere is not grounded and has to remain chargeless, by adding $-Q_{\rm ext\_im}$ uniformly distributed at the surface.

So the external potential is now known: $$ V_\text{ext}({\bf x}) = \frac{5Q/3}{4\pi\varepsilon_0|{\bf x}|} +\frac{-2Q/3}{4\pi\varepsilon_0|{\bf x}-{\bf x}_{\rm ext\_im}|} +\frac{Q}{4\pi\varepsilon_0|{\bf x}-{\bf x}_{\rm ext}|} $$ which gives the potential at (any point of) the surface of the shell as: $$ V_\text{shell} = \frac{5Q}{24\,\pi\varepsilon_0\,R} \tag{1} $$

On the inside we don't see what happens at the outside, only that the shell is at potential $V_\text{shell}$. We also have the charge $Q$ at position ${\bf x}_{\rm int}=(-R/2,0,0)$ and we give it an image charge $-2Q$ at position ${\bf x}_{\rm int\_im}=(-2R,0,0)$ so the potential in the cavity is: $$ V_\text{int}({\bf x}) = V_\text{shell}+\frac{Q}{4\pi\varepsilon_0|{\bf x}-{\bf x}_{\rm int}|} +\frac{-2Q}{4\pi\varepsilon_0|{\bf x}-{\bf x}_{\rm int\_im}|} $$ which gives the potential in the center as: $$ V_\text{cen} = \frac{11Q}{24\,\pi\varepsilon_0\,R} \tag{2} $$ NB: In the above we assume that the reader is accustomed with using [image charges] for a sphere, which always scale with the relative distance to the spherical image surface, and of course have opposite sign.

Since we are asked find the potential of the shell we must assume that it is not grounded (so it is not kept at $V=0$) and that means the total charge on the shell remains $0$. If not, the net charge on the shell would be an extra free parameter, $Q_s$, which would have to be specified before we can answer the questions.

On the outside we don't see what happens inside, but only that the total charge contained is $+Q$. We can now:

1. Distribute the internal charge $+Q$ equally over the surface, without the external $Q$ present. That gives $V=Q/(4\pi\varepsilon_0|{\bf r}|)$.
2. Subsequently put the external $Q$ at its position ${\bf r}_{\rm ext}=(3R,0,0)$ and give it the correct image charge $Q_{\rm ext\_im}=-2Q/3$ inside the sphere at location ${\bf r}_{\rm ext\_im}=(4R/3,0,0)$
3. Compensate for the image charge because the sphere is not grounded and has to remain chargeless, by adding $-Q_{\rm ext\_im}$ uniformly distributed at the surface.

So the external potential is now known: $$ V_\text{ext}({\bf r}) = \frac{5Q/3}{4\pi\varepsilon_0|{\bf r}|} +\frac{-2Q/3}{4\pi\varepsilon_0|{\bf r}-{\bf r}_{\rm ext\_im}|} +\frac{Q}{4\pi\varepsilon_0|{\bf r}-{\bf r}_{\rm ext}|} $$ which gives the potential at (any point of) the surface of the shell as: $$ V_\text{shell} = \frac{5Q}{24\,\pi\varepsilon_0\,R} \tag{1} $$

On the inside we don't see what happens at the outside, only that the shell is at potential $V_\text{shell}$. We also have the charge $Q$ at position ${\bf r}_{\rm int}=(-R/2,0,0)$ and we give it an image charge $-2Q$ at position ${\bf r}_{\rm int\_im}=(-2R,0,0)$ so the potential in the cavity is: $$ V_\text{int}({\bf r}) = V_\text{shell}+\frac{Q}{4\pi\varepsilon_0|{\bf r}-{\bf r}_{\rm int}|} +\frac{-2Q}{4\pi\varepsilon_0|{\bf r}-{\bf r}_{\rm int\_im}|} $$ which gives the potential in the center as: $$ V_\text{cen} = \frac{11Q}{24\,\pi\varepsilon_0\,R} \tag{2} $$ NB: In the above we assume that the reader is accustomed with using [image charges] for a sphere, which always scale with the relative distance to the spherical image surface, and of course have opposite sign.

Since we are asked "find the potential of the shell", we must assume that it is not grounded (so it is not kept at $V=0$) and that means the total charge on the shell remains $0$. If not, the net charge on the shell would be an extra free parameter, $Q_s$, which would have to be specified before we can answer the questions.

On the outside we don't see what happens inside, but only that the total charge contained is $+Q$. We can now:

1. Distribute the internal charge $+Q$ equally over the surface, without the external $Q$ present. That gives $V({\bf x})=Q/(4\pi\varepsilon_0|{\bf x}|)$.
2. Subsequently put the external $Q$ at its position ${\bf x}_{\rm ext}=(3R,0,0)$ and give it the correct image charge $Q_{\rm ext\_im}=-2Q/3$ inside the sphere at location ${\bf x}_{\rm ext\_im}=(4R/3,0,0)$
3. Compensate for the image charge because the sphere is not grounded and has to remain chargeless, by adding $-Q_{\rm ext\_im}$ uniformly distributed at the surface.

So the external potential is now known: $$ V_\text{ext}({\bf x}) = \frac{5Q/3}{4\pi\varepsilon_0|{\bf x}|} +\frac{-2Q/3}{4\pi\varepsilon_0|{\bf x}-{\bf x}_{\rm ext\_im}|} +\frac{Q}{4\pi\varepsilon_0|{\bf x}-{\bf x}_{\rm ext}|} $$ which gives the potential at (any point of) the surface of the shell as: $$ V_\text{shell} = \frac{5Q}{24\,\pi\varepsilon_0\,R} \tag{1} $$

On the inside we don't see what happens at the outside, only that the shell is at potential $V_\text{shell}$. We also have the charge $Q$ at position ${\bf x}_{\rm int}=(-R/2,0,0)$ and we give it an image charge $-2Q$ at position ${\bf x}_{\rm int\_im}=(-2R,0,0)$ so the potential in the cavity is: $$ V_\text{int}({\bf x}) = V_\text{shell}+\frac{Q}{4\pi\varepsilon_0|{\bf x}-{\bf x}_{\rm int}|} +\frac{-2Q}{4\pi\varepsilon_0|{\bf x}-{\bf x}_{\rm int\_im}|} $$ which gives the potential in the center as: $$ V_\text{cen} = \frac{11Q}{24\,\pi\varepsilon_0\,R} \tag{2} $$ NB: In the above we assume that the reader is accustomed with using [image charges] for a sphere, which always scale with the relative distance to the spherical image surface, and of course have opposite sign.

Since we are asked find the potential of the shell we must assume that it is not grounded (so it is not kept at $V=0$) and that means the total charge on the shell remains $0$. If not, the net charge on the shell would be an extra free parameter, $Q_s$, which would have to be specified before we can answer the questions.

On the outside we can now:

1. Distribute the internal charge $Q$ equally over the surface, without the external $Q$ present. That gives $V=Q/(4\pi\varepsilon_0|{\bf r}|)$.
2. Subsequently put the external $Q$ at its position ${\bf r}_{\rm ext}=(3R,0,0)$ and give it the correct image charge $Q_{\rm ext\_im}=-2Q/3$ inside the sphere at location ${\bf r}_{\rm ext\_im}=(4R/3,0,0)$
3. Compensate for the image charge because the sphere is not grounded and has to remain chargeless, by adding $-Q_{\rm ext\_im}$ uniformly distributed at the surface.

So the external potential is now known: $$ V_\text{ext}({\bf r}) = \frac{5Q/3}{4\pi\varepsilon_0|{\bf r}|} +\frac{-2Q/3}{4\pi\varepsilon_0|{\bf r}-{\bf r}_{\rm ext\_im}|} +\frac{Q}{4\pi\varepsilon_0|{\bf r}-{\bf r}_{\rm ext}|} $$ which gives the potential at (any point of) the surface of the shell as: $$ V_\text{shell} = \frac{5Q}{24\,\pi\varepsilon_0\,R} $$

On the inside we see nothing of what happens at the outside, we only have the charge $Q$ at position ${\bf r}_{\rm int}=(-R/2,0,0)$ and we give it an image charge $-2Q$ at position ${\bf r}_{\rm int\_im}=(-2R,0,0)$ so the potential in the cavity is: $$ V_\text{int}({\bf r}) = \frac{Q}{4\pi\varepsilon_0|{\bf r}|-{\bf r}_{\rm int}|} +\frac{-2Q}{4\pi\varepsilon_0|{\bf r}-{\bf r}_{\rm int\_im}|} $$ which gives the potential in the center as: $$ V_\text{cen} = \frac{Q}{4\,\pi\varepsilon_0\,R} $$ NB: In the above we assume that the reader is accustomed with using [image charges] for a sphere, which always scale with the relative distance to the spherical image surface, and of course have opposite sign.

Since we are asked find the potential of the shell we must assume that it is not grounded (so it is not kept at $V=0$) and that means the total charge on the shell remains $0$. If not, the net charge on the shell would be an extra free parameter, $Q_s$, which would have to be specified before we can answer the questions.

On the outside we don't see what happens inside, but only that the total charge contained is $+Q$. We can now:

1. Distribute the internal charge $+Q$ equally over the surface, without the external $Q$ present. That gives $V=Q/(4\pi\varepsilon_0|{\bf r}|)$.
2. Subsequently put the external $Q$ at its position ${\bf r}_{\rm ext}=(3R,0,0)$ and give it the correct image charge $Q_{\rm ext\_im}=-2Q/3$ inside the sphere at location ${\bf r}_{\rm ext\_im}=(4R/3,0,0)$
3. Compensate for the image charge because the sphere is not grounded and has to remain chargeless, by adding $-Q_{\rm ext\_im}$ uniformly distributed at the surface.

So the external potential is now known: $$ V_\text{ext}({\bf r}) = \frac{5Q/3}{4\pi\varepsilon_0|{\bf r}|} +\frac{-2Q/3}{4\pi\varepsilon_0|{\bf r}-{\bf r}_{\rm ext\_im}|} +\frac{Q}{4\pi\varepsilon_0|{\bf r}-{\bf r}_{\rm ext}|} $$ which gives the potential at (any point of) the surface of the shell as: $$ V_\text{shell} = \frac{5Q}{24\,\pi\varepsilon_0\,R} \tag{1} $$

On the inside we don't see what happens at the outside, only that the shell is at potential $V_\text{shell}$. We also have the charge $Q$ at position ${\bf r}_{\rm int}=(-R/2,0,0)$ and we give it an image charge $-2Q$ at position ${\bf r}_{\rm int\_im}=(-2R,0,0)$ so the potential in the cavity is: $$ V_\text{int}({\bf r}) = V_\text{shell}+\frac{Q}{4\pi\varepsilon_0|{\bf r}-{\bf r}_{\rm int}|} +\frac{-2Q}{4\pi\varepsilon_0|{\bf r}-{\bf r}_{\rm int\_im}|} $$ which gives the potential in the center as: $$ V_\text{cen} = \frac{11Q}{24\,\pi\varepsilon_0\,R} \tag{2} $$ NB: In the above we assume that the reader is accustomed with using [image charges] for a sphere, which always scale with the relative distance to the spherical image surface, and of course have opposite sign.