The total mass of the disc = \(\sigma \) \( \times \pi {R^2} = 9M\)

The mass of the circular disc cut

\( = \sigma  \times \pi {\left( {\frac{R}{3}} \right)^2} = \,\sigma  \times \frac{{\pi {R^2}}}{9} = M\)

Let us consider the above system as a complete disc of mass \(9M\) and a negative mass \(M\) super imposed on it. Moment of inertia \((I_1)\) of the complete disc = \(\frac{1}{2}9M{R^2}\) about an axis passing through \(O\) and perpendicular to the plane of the disc.

\(M.I.\) of the cut out portion about an axis passing through \(O'\) and perpendicular to the plane of disc \(=\) \(\frac{1}{2} \times M \times {\left( {\frac{R}{3}} \right)^2}\)

\(\therefore \,M.{\mathop{\rm I}\nolimits} .\left( {{I_2}} \right)\) of the cut out portion about an axis passing through \(O\) and perpendicular to the plane of disc

\( = \left[ {\frac{1}{2} \times M \times {{\left( {\frac{R}{3}} \right)}^2} + M \times {{\left( {\frac{{2R}}{3}} \right)}^2}} \right]\)

                                                  [Using perpendicular axis theorem]

\(\therefore \) The total \(M.I.\) of the system about an axis passing through \(O\) and perpendicular to the plane of the disc is 

\(\begin{array}{l}
 = \left[ {\frac{1}{2} \times M \times {{\left( {\frac{R}{3}} \right)}^2} + M \times {{\left( {\frac{{2R}}{3}} \right)}^2}} \right]\\
I = {I_1} + {I_2}\\
 = \frac{1}{2}9M{R^2} - \left[ {\frac{1}{2} \times M \times {{\left( {\frac{R}{3}} \right)}^2} + M \times {{\left( {\frac{{2R}}{3}} \right)}^2}} \right]\\
 = \frac{{9M{R^2}}}{2} - \frac{{9M{R^2}}}{{18}} = \frac{{\left( {9 - 1} \right)M{R^2}}}{2} = 4M{R^2}
\end{array}\)