Let, mass of entire disc $=M$
Mass per unit area $=\frac{M}{\pi(2 R)^2}=\frac{M}{4 \pi R^2}$
Mass of removed disc of radius $R$,
$$
M_1=\frac{M}{4 \pi R^2} \times R^2=\frac{\mathrm{M}}{4}
$$

Mass of remaining disc,
$$
\Rightarrow \quad M_2=M-\frac{M}{4}=\frac{3 M}{4}
$$

Center of mass of removed disc is $C_1$ and centre of mass of remaining new disc is $C_2$.

And centre of mass of combination of $M_1$ and $M_2$ will be at $C(0,0)$.
$$
\begin{aligned}
\Rightarrow \quad & \frac{M_1 x_1-M_2 x_2}{M_1-M_2}=0 \\
\Rightarrow \quad & M_1 x_1=M_2 x_2 \\
& \frac{M}{4} \cdot R=\frac{3 M}{4}(\alpha R) \quad \Rightarrow \alpha=\frac{1}{3}
\end{aligned}
$$