The moment of inertia of a uniform circular disc about an axis perpendicular to its plane and passing through its centre is given by $I = \frac{1}{2}MR^2$.

For the original complete disc: Mass, $M_1 = 9M$ Radius, $R_1 = R$ Moment of inertia, $I_1 = \frac{1}{2} M_1 R_1^2 = \frac{1}{2}(9M)R^2 = \frac{9}{2}MR^2$

For the small disc that is removed: Mass, $M_2 = M$ Radius, $R_2 = R/3$ Moment of inertia, $I_2 = \frac{1}{2} M_2 R_2^2 = \frac{1}{2}(M)\left(\frac{R}{3}\right)^2 = \frac{1}{2}M\left(\frac{R^2}{9}\right) = \frac{1}{18}MR^2$

According to the principle of superposition, the moment of inertia of the remaining disc is the difference between the moment of inertia of the original disc and the removed disc: $I_{remaining} = I_1 - I_2$ $I_{remaining} = \frac{9}{2}MR^2 - \frac{1}{18}MR^2$ $I_{remaining} = \left(\frac{81 - 1}{18}\right)MR^2$ $I_{remaining} = \frac{80}{18}MR^2 = \frac{40}{9}MR^2$