There is a charge at the center with charge, $q$, and a charged shell with finite width with inner and outer radii $a$ and $b$ respectively and charge density $\rho$, if I understand correctly.
Then there are three regions: $(0,a)\cup(a,b)\cup(b,\infty)$ combined that we have to find the field in and then plot. Using Gauss' law and focusing on the inner most region we have
$$\oint E_i da_i = \frac{q_{\text{enc}}}{\epsilon_0}$$
Then because of the spherical symmetry we can write
\begin{equation}
E4\pi r^2=\frac{q}{\epsilon_0}\implies E_r=\frac{q}{4\pi\epsilon_0 r^2}
\end{equation}
Then for the middle region we can do the same thing but a little more explicit
$$\oint E_i da_i =\frac{1}{\epsilon_0}\int \rho\, d\tau$$
For this region then we have
$$E4\pi r'^2=\frac{\rho}{\epsilon_0}\int_{a}^{r'}d\tau =\frac{\rho}{\epsilon_0}\frac{4}{3}\pi (r'^3 - a^3)\implies E_r=\frac{\rho (r'^3 -a^3)}{3\epsilon_0 r'^2}$$
Then totally in that region we have
$$E_r=\frac{\rho (r'^3 -a^3)}{3\epsilon_0 r'^2}+\frac{q}{4\pi\epsilon_0 r'^2}$$
Finally for completely outside the shell we simply note that the total charge is the charge inside, and the whole shell, which really means let $r'$ in the first term go to $b$, i.e.
$$E_r=\frac{\rho (b^3 -a^3)}{3\epsilon_0 r'^2}+\frac{q}{4\pi\epsilon_0 r'^2}$$
since we don't get any more charge outside of the shell's radius.
