The mass density of planet of radius R varies with the distance r from its centre as $\mathrm{\rho }\left(\mathrm{r}\right)={\mathrm{\rho }}_0\left(1-\frac{{\displaystyle {\mathrm{r}}^2}}{{\displaystyle 2{\mathrm{R}}^2}}\right)$. Then the gravitational field is maximum at:

Mass of small element of planet of radius x and thickness dx is
$\mathrm{dm}=\mathrm{\rho }\times 4{\mathrm{\pi x}}^2\mathrm{dx}={\mathrm{\rho }}_0(1-\frac{{\mathrm{x}}^2}{2{\mathrm{R}}^2})\times 4{\mathrm{\pi x}}^2\mathrm{dx}$
Mass of the planet
$$\begin{gathered} \mathrm{M}=4{\mathrm{\pi \rho }}_0\underset{\mathrm{O}}{\overset{\mathrm{r}}{\int }}({\mathrm{x}}^2-\frac{{\mathrm{x}}^4}{2{\mathrm{R}}^2})\mathrm{dx} \\ \Rightarrow \mathrm{M}=4{\mathrm{\pi \rho }}_0\left|\frac{{\displaystyle {\mathrm{r}}^3}}{{\displaystyle 3}}-\frac{{\displaystyle {\mathrm{r}}^5}}{{\displaystyle 10{\mathrm{R}}^2}}\right| \end{gathered}$$
Gravitational field $\mathrm{E}=\frac{\mathrm{GM}}{{\mathrm{r}}^2}=\frac{\mathrm{G}}{{\mathrm{r}}^2}\times 4{\mathrm{\pi \rho }}_0\left(\frac{{\mathrm{r}}^3}{3}-\frac{{\mathrm{r}}^5}{10{\mathrm{R}}^2}\right)$
$\Rightarrow \mathrm{E}=4{\mathrm{\pi G\rho }}_0\left(\frac{{\displaystyle \mathrm{r}}}{{\displaystyle 3}}-\frac{{\displaystyle {\mathrm{r}}^3}}{{\displaystyle 10{\mathrm{R}}^2}}\right)$
E is maximum when $\frac{\mathrm{dE}}{\mathrm{dr}}=0$
$$\begin{gathered} \Rightarrow \frac{\mathrm{dE}}{\mathrm{dr}}=4{\mathrm{\pi G\rho }}_0(\frac{1}{3}-\frac{3{\mathrm{r}}^2}{10{\mathrm{R}}^2})=0 \\ \Rightarrow \mathrm{r}=\frac{\sqrt{10}}{3}\mathrm{R} \end{gathered}$$