Consider a spherical shell of radius $\mathrm{r}$ and radial thickness dr. $P \& P+d P$ are pressure at its inner and outer surface.

Let $\mathrm{g}_{\mathrm{r}}=$ gravitational acceleration at distance $\mathrm{r}( < \mathrm{R})$
$\sum_{\downarrow} \mathrm{P}_{0}=$ Atmospheric pressure

For equilibrium of this shell -
$$
\begin{array}{l}
(P+d P)\left(4 \pi r^{2}\right)+\left(4 \pi r^{2} d r\right) \rho g_{r}=(P)\left(4 \pi r^{2}\right) \\
\left\{\rho=\frac{3 M}{4 \pi R^{3}}=\right.\text { Density of sphere } \\
\Rightarrow \quad d P=-\rho g_{r} d r \\
\because \quad g_{r}=\frac{4}{3} \pi \text { Gpr } \\
\Rightarrow \quad \mathrm{dP}=-\frac{4}{3} \pi \mathrm{G} \rho^{2} r d r
\end{array}
$$
$\{(-)$ ve sign indicates that pressure is decreasing with radius $\}$
$$
\begin{array}{l}
\Rightarrow \int_{P}^{P_{0}} d P=\int_{r}^{R}-\frac{4}{3} \pi G \rho^{2} r d r \\
\Rightarrow & P_{0}-P=-\frac{4}{3} \pi G \rho^{2}\left(\frac{R^{2}}{2}-\frac{r^{2}}{2}\right) \\
\Rightarrow & P=P_{0}+\frac{4}{3} \pi G \rho^{2}\left(\frac{R^{2}}{2}-\frac{r^{2}}{2}\right) \\
& P=P_{0}+\frac{3 G M^{2}}{8 \pi R^{4}}\left(1-\frac{r^{2}}{R^{2}}\right)
\end{array}
$$
Hence $P \propto \frac{1}{\mathrm{R}^{4}}$
$\Rightarrow$ Average pressure will also be proportional to $\frac{1}{\mathrm{R}^{4}}$
$\Rightarrow$ correct answer is $[\mathrm{A}]$