A disk of certain radius is cut from a disk of mass 9M and radius R as shown in figure. Find its moment of inertia about an axis passing through its centre C and perpendicular to its plane.

Moment of inertia of given disk (before cutting) (w.r.t axis passing through it's centre C and perpendicular to it's plane) or (principle axis) = $\frac{\left(9\mathrm{M}\right){\mathrm{R}}^2}{2}$
Mass of given disk = 9M
(Assume density and thickness are uniform every where on disk)
So mass for area π(R²) = 9M
So mass per unit area = $=\frac{9\mathrm{M}}{\pi {\mathrm{R}}^2}$
So mass for area
$\pi {\left(\frac{\mathrm{R}}{3}\right)}^2=\frac{9\mathrm{M}}{\pi {\mathrm{R}}^2}\times \pi {\left(\frac{\mathrm{R}}{3}\right)}^2$
= M = mass of small disk

So moment of inertia of small disk (w.r.t. its own principle axis) = $\frac{\mathrm{M}{\left(\frac{\mathrm{R}}{3}\right)}^2}{2}$

Now from parallel axis theorem

$\mathrm{I}={\mathrm{I}}_{\mathrm{cm}}+{\mathrm{Md}}^2$

here d is distance between parallel axes,

So ${\mathrm{I}}_c=\frac{\mathrm{M}{\left(\frac{\mathrm{R}}{3}\right)}^2}{2}+\mathrm{M}{\left(\frac{2\mathrm{R}}{3}\right)}^2$

moment of inertia of small disk wrt principle axis of bigger disk)

(here, $\mathrm{d}=\frac{2\mathrm{R}}{3},$ given in question)

Solving this ${\mathrm{I}}_c=\frac{{\mathrm{MR}}^2}{18}+\frac{4{\mathrm{MR}}^2}{9}$

${\mathrm{I}}_c=\frac{{\mathrm{MR}}^2}{2}$

Now moment of inertia of remaining part w.r.t. given axis

$=\frac{9{\mathrm{MR}}^2}{2}-\frac{{\mathrm{MR}}^2}{2}=4{\mathrm{MR}}^2$