Let the moment of inertia of about the centre of the large disc:

the large (unpunctured) disc be $\displaystyle I_A$ ;

the small cut out disc be $\displaystyle I_B$;

the punctured large disc be $\displaystyle I_{AB}$.

Then we must have:

$\displaystyle I_{AB} + I_B = I_A$

where

$\displaystyle I_A = \frac{1}{2} M R^2$

and

$\displaystyle I_B = \frac{1}{2}m r^2 + m x^2$ (Parallel axis theorem)

so

$\displaystyle I_{AB} = I_A - I_B = \frac{1}{2}M R^2 - \frac{1}{2} m r^2 - m x^2$

but

$\displaystyle m = \frac{r^2}{R^2}M$ (same density and similarity of shape)

so

$\displaystyle \color[rgb]{0,0,1}\boxed{I_{AB} = \frac{1}{2}\left(R^2 - \frac{r^4}{R^2} - \frac{2 r^2 x^2}{R^2} \right)M}$