A circular disc of radius $R$ is removed from a bigger circular disc of radius $2R$ such that the circumferences of the discs coincide. The centre of mass of the new disc is $\alpha R$ from the centre of the bigger disc. The value of $\alpha$ is:

Let the center of the large disc be at the origin $(0,0)$.The area of the large disc is $A_0=\pi (2R{)}^2=4\pi R^2$.The removed disc has radius $R$ and its center is at $(R,0)$ to make circumferences coincide at $(2R,0)$.Area of the removed disc is $A_1=\pi R^2$.Using the center of mass formula for a composite body with a hole:
$$x_{cm}=\frac{A_0{x}_0-A_1{x}_1}{A_0-A_1}$$
$$x_{cm}=\frac{(4\pi R^2)(0)-(\pi R^2)(R)}{4\pi R^2-\pi R^2}$$
$$x_{cm}=\frac{-\pi R^3}{3\pi R^2}=-\frac{R}{3}$$
The distance from the center is $|x_{cm}|=R/3$.Comparing with $\alpha R$, we get $\alpha =1/3$.