A thin disc of mass M and radius R has mass per unit area $$\sigma(r) = kr^2$$ where r is the distance from its centre. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is:

We begin by recalling that for any surface with surface mass density $$\sigma(r)$$, an infinitesimal ring of radius $$r$$ and width $$dr$$ has an area $$dA = 2\pi r\,dr$$ (because the circumference is $$2\pi r$$ and the thickness is $$dr$$). The small mass of this ring is therefore $$dm = \sigma(r)\,dA$$.

The given surface mass density for the disc is $$\sigma(r) = k r^{2}$$, where $$k$$ is a constant that we shall determine from the total mass $$M$$ of the disc.

First, we write the expression for the total mass. By definition,

$$M = \int_{0}^{R} dm = \int_{0}^{R} \sigma(r)\,dA.$$

Substituting $$\sigma(r) = k r^{2}$$ and $$dA = 2\pi r\,dr$$, we get

$$M = \int_{0}^{R} k r^{2}\,(2\pi r\,dr) = 2\pi k \int_{0}^{R} r^{3}\,dr.$$

We now integrate $$r^{3}$$:

$$\int r^{3}\,dr = \frac{r^{4}}{4}.$$ Hence

$$M = 2\pi k\left[\frac{r^{4}}{4}\right]_{0}^{R} = 2\pi k\left(\frac{R^{4}}{4}\right) = \frac{\pi k R^{4}}{2}.$$

Solving this relation for the constant $$k$$ gives

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

Now we calculate the moment of inertia about the central axis (perpendicular to the plane). The general formula for a surface is

$$I = \int r^{2}\,dm,$$ where again $$dm = \sigma(r)\,dA.$$ Substituting both $$\sigma(r)$$ and $$dA$$, we obtain

$$I = \int_{0}^{R} r^{2}\,\bigl(k r^{2}\bigr)\,(2\pi r\,dr) = 2\pi k \int_{0}^{R} r^{5}\,dr.$$

We now integrate $$r^{5}$$: $$\int r^{5}\,dr = \frac{r^{6}}{6}.$$ Therefore

$$I = 2\pi k\left[\frac{r^{6}}{6}\right]_{0}^{R} = 2\pi k\left(\frac{R^{6}}{6}\right) = \frac{\pi k R^{6}}{3}.$$

Substituting the value of $$k = \dfrac{2M}{\pi R^{4}}$$ that we found earlier, we get

$$I = \frac{\pi}{3}\left(\frac{2M}{\pi R^{4}}\right)R^{6} = \frac{2M R^{2}}{3}.$$

This simplifies neatly to

$$I = \frac{2 M R^{2}}{3}.$$

Hence, the correct answer is Option D.