Interesting question! Let's try to answer it.
Magnetic confinement fusion is at the moment the most successful attempt trying to build a fusion reactor on Earth. You mentioned ITER which belongs to this class. ITER will have a plasma volume of $840\,\mathrm{m}^3$ and a fusion power of $500\,\mathrm{MW}$, resulting in a power density of roughly $600\,\mathrm{kW}/\mathrm{m}^{3}$. You are asking if it is possible to have a self-sustained fusion power-plant in the megawatt-range with the same power density as the sun. I guess that you have a typo there and wanted to write gigawatt-range instead (I am assuming $1\,\mathrm{GW}$ here).
Let's first estimate the volume of your plasma:
\begin{equation}
V = \frac{1\,\mathrm{GW}}{276.5\,\mathrm{W}/\mathrm{m}^3}
\approx 3.6\cdot10^6\,\mathrm{m}^3.
\end{equation}
This corresponds to a sphere with a radius of $r\approx95\,\mathrm{m}$ or a torus with a large radius of $R\approx118\,\mathrm{m}$ (assuming a typical aspect ratio of $A=3$).
Let's now estimate the number of fusion reactions required to keep your plasma burning for a time of $1\,\mathrm{s}$. The energy released over that time period amounts to
\begin{equation}
W = P\cdot \Delta t = 1\,\mathrm{GW}\cdot1\mathrm{s} = 1\,\mathrm{GJ}.
\end{equation}
The energy released in a single fusion reaction can be calculated from the mass difference of the fusion product compared to its initial ingredients. The mass difference is for D-T fusion $\Delta m=0.018835\,\mathrm{u}$, with $\mathrm{u}$ the atomic mass unit, which corresponds to an energy of $\Delta E\approx2.81\,\mathrm{nJ}$. We can now estimate the number of fusion reactions to
\begin{equation}
N = \frac{1\,\mathrm{GJ}}{\Delta E}
\approx 3.6 \cdot 10^{20}.
\end{equation}
Next, we will have a look at the fusion reaction rate coefficient $\mathcal{R}$, which is the number of reactions per volume per time. I am assuming D-T fusion here (and not p-p fusion which is mostly what happens in our sun). $\mathcal{R}$ is defined as
\begin{equation}
\mathcal{R}
= n_\mathrm{deuterium} n_\mathrm{tritium} \left<\sigma v \right>
= \left(\frac{n}{2}\right)^2\left<\sigma v \right>,
\end{equation}
with $\left<\sigma v \right>$ the fusion reactivity (depending on the cross section $\sigma$ and the relative velocity $v$ of the two fusing particles). Since $\mathcal{R}$ corresponds to the number of reactions per volume per time, we can calculate it as
\begin{equation}
\mathcal{R}
= \frac{N}{V\cdot\Delta t}
\approx \frac{3.6\cdot10^{20}}{3.6\cdot10^6\,\mathrm{m}^3\cdot 1\,\mathrm{s}}
= 10^{14}\,\frac{1}{\mathrm{m}^3\mathrm{s}}
\end{equation}
Looking at the fusion reactivity will now allow us to estimate the plasma density. Using a wikipedia article as reference, we have a close look at the plot of the fusion reactivity (plotted as a function of the plasma temperature). We see that the DT reactivity has a maximum at a temperature of around $50\,\mathrm{keV}$. The maximum value is roughly
\begin{equation}
\left<\sigma v \right>_\mathrm{max}
\approx 10^{-21}\,\mathrm{m}^3/\mathrm{s}.
\end{equation}
Rearranging the equation for the reactivity, we can estimate the plasma density:
\begin{equation}
n = \sqrt{ 2\frac{\mathcal{R}}{\left<\sigma v \right>_\mathrm{max}} }
\approx \sqrt{ 2 \frac{10^{14}\,\frac{1}{\mathrm{m}^3\mathrm{s}}}{10^{-21}\,\mathrm{m}^3/\mathrm{s}} }
= \sqrt{ 2\cdot 10^{35}\,\mathrm{m}^{-6} }
\approx 4.5\cdot10^{17}\,\mathrm{m}^{-3}.
\end{equation}
Finally, we apply the Lawson criterion (or the triple product), again a link to the corresponding wikipedia article, which let's us estimate the required energy confinement time (since we have set plasma density $n$ and temperature $T$ already) to get a self-sustained plasma:
\begin{equation}
n \cdot T \cdot \tau_E > 3\cdot 10^{24}\,\frac{\mathrm{eV}\,\mathrm{s}}{\mathrm{m}^3}.
\end{equation}
Rearranging and inserting the values for $n$ and $T$ yields
\begin{equation}
\tau_E > \frac{ 3\cdot 10^{24}\,\frac{\mathrm{eV}\,\mathrm{s}}{\mathrm{m}^3} }
{4.5\cdot10^{17}\,\mathrm{m}^{-3} \cdot 50\,\mathrm{keV}}
\approx 130\,\mathrm{s}.
\end{equation}
Note that this value is two orders of magnitude above what is currently possible. Apart from that, you now know what should be required to operate a fusion power plant with the same power density as the sun.
