From a circular disc of radius R and mass 9 M, a small disc or radius R/3 is removed. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is

Mass per unit area of disc = $\frac{9M}{\pi R^2}$

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

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

$I_1=\frac{M}{2}{\left(\frac{R}{3}\right)}^2+M{\left(\frac{2R}{3}\right)}^2=\frac{1}{2}M{R}^2$

When portion of disc would not have been removed, then the moment of inertia of complete disc about the given axis would be

$I_2=\frac{1}{2}M{R}^2$

So moment of inertia of the disc with removed portion, about the given axis is:

$I=I_2-I_1=\frac{9}{2}M{R}^2-\frac{1}{2}M{R}^2=4M{R}^2$