Variation in Density!

As the density is having spherical symmetry, let us take an element as an infinitesimally thin hollow sphere concentric with the main sphere with radius $r$ and thickness $dr$ .

$\therefore dI = \dfrac{2}{3}r^2dm \hspace{3 cm} \Bigg(\because I_{\text{hollow sphere}} = \dfrac{2}{3}MR^2\Bigg)$

For the element, $dm = \rho \cdot dV = \rho \cdot 4\pi r^2dr$

$\therefore dI = \dfrac{2}{3}r^2\cdot\rho \cdot 4\pi r^2dr$

$\implies dI = \dfrac{8\pi}{3} \rho r^4 dr$

$\implies I = \dfrac{8\pi}{3\cdot 5} \displaystyle\int_{0}^{R} \rho\cdot 5r^4 dr$

$\implies I = \dfrac{8\pi}{3\cdot 5} \displaystyle\int_{0}^{R} \rho\cdot d(r^5)$

The above integral is simply the area under the graph of $\rho$ vs $r^5$ which can be easily calculated from the above graph as $\frac{1}{2}\cdot\rho\cdot R^5$

$\therefore I = \boxed{ \dfrac{4}{15}\pi \rho_0 R^5 }$