A disc of radius R is cut out from a larger disc of radius 2R such...

Let the centre of the larger disc be at the origin $(0,0)$.

Radius of the larger disc is $2R$. Its mass is proportional to its area, $M_1 = \sigma \pi (2R)^2 = 4\sigma \pi R^2$, where $\sigma$ is the surface mass density.

The hole has a radius $R$ and touches the edge of the larger disc. The distance of its centre from the origin is $2R - R = R$. Let its centre be at $(R, 0)$.

Mass of the cut-out disc is $M_2 = \sigma \pi R^2$.

The $x$-coordinate of the centre of mass of the residual disc is given by:

$x_{cm} = \dfrac{M_1 x_1 - M_2 x_2}{M_1 - M_2}$

Substituting the values:

$x_{cm} = \dfrac{4\sigma \pi R^2 (0) - \sigma \pi R^2 (R)}{4\sigma \pi R^2 - \sigma \pi R^2}$

$x_{cm} = \dfrac{-\sigma \pi R^3}{3\sigma \pi R^2} = -\dfrac{R}{3}$

The negative sign indicates that the centre of mass of the residual disc is at a distance of $R/3$ from the centre of the original disc, on the side opposite to the hole.