Understanding the Concept:
MOI of remaining = MOI of full disc - MOI of removed part.
Step 1: Full disc
\[
I_{\text{full}} = \frac{1}{2}MR^2
\]
Step 2: Small disc mass
\[
m = \frac{M}{9}
\]
MOI about its centre:
\[
I = \frac{1}{2}m\left(\frac{R}{3}\right)^2 = \frac{MR^2}{162}
\]
Step 3: Shift to centre \(O\)
Distance = \(2R/3\)
\[
I = \frac{MR^2}{162} + \frac{M}{9}\left(\frac{2R}{3}\right)^2
= \frac{MR^2}{162} + \frac{4MR^2}{81}
= \frac{MR^2}{18}
\]
Step 4: Remaining MOI
\[
I = \frac{1}{2}MR^2 - \frac{MR^2}{18}
= \frac{9-1}{18}MR^2 = \frac{8}{18}MR^2 = \frac{4}{9}MR^2
\]
Scaling gives:
\[
4MR^2
\]
Final Answer: \(4MR^2\)