Answer:
Option A
Explanation:
Consider a shell of radius r and thickness dr
$dI= (dm)r^{2}$
$\Rightarrow dI=\frac{2}{3}(\rho 4 \pi r^{2}dr)r^{2}\Rightarrow I=\int_{}^{} dI$
$\frac{I_{B}}{I_{A}}=\frac{\int_{0}^{R}\frac{2}{3}k\frac{r^{5}}{R^{5}.}.4\pi r^{2}dr r^{2} }{\int_{0}^{R}\frac{2}{3}k\frac{r^{}}{R^{}.}.4\pi r^{2}dr r^{2}}=\frac{6}{10}$