Ans this aLsO

Dear Student
$$\begin{gathered} Let mass per unit area of the original disc=\sigma \\ Thus mass of original disc=M=\sigma A =\sigma \pi R^2 \\ Radius of smaller disc=R/2. \\ Thus mass of the smaller disc=\sigma \pi {(R/2)}^2 =M/4 \\ Centre of mass \\ {x}_{cm} =\frac{{\displaystyle \sum _{}}{M}_i{x}_i}{\sum _{}{M}_i}, y_{cm} =\frac{{\displaystyle \sum _{}{M}_i{y}_i}}{\sum _{}Mi} \\ Here \\ {x}_{i } is x coordinates of centre of masses \\ Let\text{'}s assume centre of mass coordinate for big disk is (0,0) \\ And that if small disk in reference of big disk =(R/2,0) \\ {x}_{cm} =\frac{M\times 0-{\displaystyle \frac{M}{4}} \times {\displaystyle \frac{R}{2}}}{M-\frac{M}{4}} (Here - sign is taken because small disk is taken out) \\ {x}_{cm} =-\frac{M}{4} \times \frac{R}{2}\times \frac{4}{3M} =-\frac{R}{6} \\ {y}_{cm} =\frac{M\times 0-{\displaystyle \frac{M}{4}} \times {\displaystyle 0}}{M-\frac{M}{4}} =0 \\ So \\ Coordinate of centre of mass (-\frac{R}{6},0) \end{gathered}$$
The negative sign indicates that the centre of mass gets shifted toward the left of point O
So correct option is B

Regards