Moment of Inertia of the Full Disc:

The moment of inertia (M.I.) of the entire disc without any cavity is given by:

$ I_1 = \frac{1}{2} MR^2 $

Mass of the Removed Disc:

The mass of the removed disc, which is of radius $ \frac{R}{3} $, is calculated as:

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

Moment of Inertia of the Removed Disc:

The moment of inertia of the removed disc involves two parts: about its center and due to its position.

About its center:

$ \frac{\frac{M}{9} \left(\frac{R}{3}\right)^2}{2} = \frac{MR^2}{162} $

Due to its position (distance from O to the new center of the small disc):

The distance is $ \frac{2R}{3} $, so:

$ \frac{M}{9} \times \left(\frac{2R}{3}\right)^2 = \frac{4MR^2}{81} $

Total M.I. of removed disc:

$ I_2 = \frac{MR^2}{162} + \frac{4MR^2}{81} = \frac{MR^2}{18} $

Moment of Inertia of the Remaining Part:

Subtract the moment of inertia of the removed disc from the full disc:

$ I = I_1 - I_2 = \frac{1}{2} MR^2 - \frac{MR^2}{18} = \frac{9MR^2 - MR^2}{18} = \frac{8MR^2}{18} = \frac{4MR^2}{9} $

Thus, the value of $x$ is $9$.