Let the unit area of the original disc be $\sigma$

The radius of the original disc is given as $2r$

Mass of the original disc can be calculated as $m=\pi\left(2 r^{2}\right) \sigma=4 \pi r^{2} \sigma \ldots \ldots \ldots \ldots \ldots$ (i)

Following figure shows the disc with the cut portion:

Radius of the smaller disc is given as $r$

Mass of the smaller disc can be calculated as $m^{\prime}=\pi r^{2} \sigma$

$\Rightarrow m^{\prime}=m / 4$

[ From equation (i) ]

Let the respective centers of the disc cut off from the original and the original disc be $\mathrm{O}$ ‘ and $\mathrm{O}$.

According to the concept of centre of mass, the original disc’s centre of mass is concentrated at O, while the smaller disc’s centre of mass is assumed to be at O’.

We know that:

$\mathrm{OO}^{\prime}=\mathrm{R} / 2=\mathrm{r}_{2}$

After the smaller circle has been cut out, the masses of the left two systems are:

$-m^{\prime}(=m / 4)$ concentrated at $O^{\prime}$, and $m$ (concentrated at O).

Let $X$ be the distance of the center of mass from $O.$

We know :

$X=\frac{{{m}_{1}}{{r}_{1}}+{{m}_{2}}{{r}_{2}}}{{{m}_{1}}+{{m}_{2}}}$

$X=\frac{[m\times 0-{{m}^{‘}}\times (\frac{r}{2})]}{M+(-M)}$

$X=\frac{-R}{6}$