1. Given:

   • Original disc has mass $M$ and radius $R$.
   • A small disc of radius $R/3$ is removed.
   • Axis $AB$ passes through the centre $O$ of the original big disc and is perpendicular to the plane.

2. Mass of the removed small disc

   Since the surface mass density is uniform, $\sigma = \frac{M}{\pi R^2}$

   Radius of removed disc $= R/3$, so its mass is $m = \sigma \cdot \pi \left(\frac{R}{3}\right)^2 = \frac{M}{\pi R^2}\cdot \pi \frac{R^2}{9} = \frac{M}{9}$

3. Position of the removed disc

   From the standard figure for this problem, the small disc is cut such that it is internally tangent to the outer boundary. Hence the distance between the centres is $d = R - \frac{R}{3} = \frac{2R}{3}$

4. Moment of inertia of the original full disc about axis through $O$

   $I_{\text{full}} = \frac{1}{2}MR^2$

5. Moment of inertia of the removed disc about the same axis

   First, about its own centre: $I_{\text{small, center}} = \frac{1}{2}m\left(\frac{R}{3}\right)^2$

   Using parallel axis theorem to shift to axis through $O$: $I_{\text{removed about }O} = \frac{1}{2}m\left(\frac{R}{3}\right)^2 + md^2$

   Substitute $m=\frac{M}{9}$ and $d=\frac{2R}{3}$: $I_{\text{removed about }O} = \frac{1}{2}\cdot \frac{M}{9}\cdot \frac{R^2}{9} + \frac{M}{9}\cdot \left(\frac{2R}{3}\right)^2$

   $= \frac{MR^2}{162} + \frac{M}{9}\cdot \frac{4R^2}{9}$

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

   $= \frac{MR^2}{162} + \frac{8MR^2}{162} = \frac{9MR^2}{162} = \frac{MR^2}{18}$

6. Moment of inertia of remaining part

   $I_{\text{remaining}} = I_{\text{full}} - I_{\text{removed about }O}$

   $= \frac{1}{2}MR^2 - \frac{1}{18}MR^2$

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

7. Compare with given form

   Given, $I_{\text{remaining}} = \frac{4}{x}MR^2$

   So, $\frac{4}{x} = \frac{4}{9} \Rightarrow x=9$

Final answer: $x=9$