Consider a disc of radius R and mass M . A hole of radius R 3 is created in the disk such that the center of…

Mass per unit area of the disc$=\frac{M}{\pi R^2}$.

Mass of removed portion of disc, $M^{\text{'}}=\frac{M}{\pi R^2}\times \pi {\left(\frac{R}{3}\right)}^2=\frac{M}{9}$

Moment of inertia of removed portion about an axis passing through centre of disc $O$ and perpendicular to the plane of disc using parallel axis theorem is 

$I_O^{\text{'}}=I_{COM}+M^{\text{'}}{d}^2$

$$\begin{gathered} =\frac{1}{2}\times \frac{M}{9}{\left(\frac{R}{3}\right)}^2+\frac{M}{9}{\left(\frac{R}{3}\right)}^2 \\ =\frac{M{R}^2}{162}+\frac{M{R}^2}{81}=\frac{3M{R}^2}{162} \end{gathered}$$

When portion of disc would not have been removed, the moment of inertia of complete disc about centre $O$ is $I_O=\frac{M{R}^2}{2}$.

So, moment of inertia of the disc with removed portion is $I=I_O-I_O^{\text{'}}=\frac{M{R}^2}{2}-\frac{3M{R}^2}{162}=\frac{78M{R}^2}{162}=\frac{13M{R}^2}{27}$