Given surface mass density

The disc has variable surface density $\sigma(r) = k r^2$ where $r$ is the distance from the centre.

Find total mass $M$

Consider a thin ring of radius $r$ and thickness $dr$.

Its area is $dA = 2\pi r\,dr$

So its mass is $dm = \sigma(r)\, dA = kr^2(2\pi r\,dr) = 2\pi k r^3\,dr$

Hence total mass: $M = \int_0^R dm = \int_0^R 2\pi k r^3\,dr$ $M = 2\pi k \left[\frac{r^4}{4}\right]_0^R = \frac{\pi k R^4}{2}$

Therefore, $k = \frac{2M}{\pi R^4}$

Find moment of inertia about the central perpendicular axis

For the same ring element, $dI = r^2\,dm$

Thus, $dI = r^2(2\pi k r^3\,dr) = 2\pi k r^5\,dr$

Integrating from $0$ to $R$, $I = \int_0^R 2\pi k r^5\,dr$ $I = 2\pi k \left[\frac{r^6}{6}\right]_0^R = \frac{\pi k R^6}{3}$

Substitute $k$ in terms of $M$

$I = \frac{\pi R^6}{3} \cdot \frac{2M}{\pi R^4}$ $I = \frac{2MR^2}{3}$

Match with options

$I = \frac{2MR^2}{3}$

So the correct option is C.