A thin disc of mass 9M and radius R from which a disc of radius R/3 is cut, is shown in figure. Then moment of inertia of the remaining disc about O, perpendicular to the plane of disc is :
 

As the mass is uniformly distributed on the disc, so mass density (per unit area) $=9\mathrm{M}/{\mathrm{\pi R}}^2$
Mass of removed portion$=\frac{9\mathrm{M\pi }}{{\mathrm{\pi R}}^2}\times {\left(\frac{\mathrm{R}}{3}\right)}^2=\mathrm{M}$
So, the moment of inertia of the removed portion about the stated axis by theorem of parallel axis is ${\mathrm{I}}_1=\frac{\mathrm{M}}{2}{\left(\frac{\mathrm{R}}{3}\right)}^2+\mathrm{M}{\left(\frac{2\mathrm{R}}{3}\right)}^2$      ....(i)
If the disc would not have been removed, then the moment of inertia of complete disc about the stated axis is${\mathrm{I}}_2=9\mathrm{M}\frac{{\mathrm{R}}^2}{2}$    .....(ii)
So, the moment of inertia of the disc shown is ${\mathrm{I}}_2-{\mathrm{I}}_1,\mathrm{i},\mathrm{e},{\mathrm{I}}_2-{\mathrm{I}}_1=4{\mathrm{MR}}^2$