A circular hole of radius R/4 is made in a thin uniform disc having mass M and radius R, as shown in figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point O and perpendicular to the plane of the disc is :

Let the disc's mass per unit area be $\sigma$.
now, removed mass's moment of inertia about the origin is

Using the parallel axis theorem,

${\mathrm{M}}_2=\left(\frac{(\mathrm{\sigma A}\text{'}){\left({\displaystyle \frac{\mathrm{R}}{4}}\right)}^2}{2}+(\mathrm{\sigma A}\text{'}){\left(\frac{3\mathrm{R}}{4}\right)}^2\right)$

${\mathrm{M}}_2=\left(\mathrm{\sigma A}\text{'}\right)\left(\frac{{\mathrm{R}}^2}{32}+\frac{9{\mathrm{R}}^2}{16}\right)$

$\mathrm{A}\text{'}=\mathrm{\pi }{\left(\frac{\mathrm{R}}{4}\right)}^2=\frac{{\mathrm{\pi R}}^2}{16}$

${\mathrm{M}}_2=\left(\frac{{\mathrm{\sigma \pi R}}^4}{16}\right)\left(\frac{19}{32}\right)$

The moment of inertia of the complete disc,

${\mathrm{M}}_1=\left(\frac{\left(\mathrm{\sigma A}\right)\times {\mathrm{R}}^2}{2}\right)$

Effect moment of inertia,

$={\mathrm{M}}_1-{\mathrm{M}}_2$

${\mathrm{M}}_{\mathrm{o}}=\frac{\mathrm{\sigma }}{2}{\mathrm{\pi R}}^4-\frac{{\mathrm{\sigma \pi R}}^{4}19}{512}=\left({\mathrm{\sigma \pi R}}^2\right){\mathrm{R}}^2\frac{237}{512}$

${\mathrm{\sigma \pi R}}^2=\mathrm{M}$

${\mathrm{M}}_{\mathrm{o}}=\frac{237{\mathrm{MR}}^2}{512}$

Hence the correct answer is $\frac{237{\mathrm{MR}}^2}{512}.$