We can consider the remaining disc as a combination of a larger disc of radius 28cm and a smaller disc of negative mass of radius 21cm.

Assuming that the linear mass density of the disc as $\rho$

$\text{mass of original disc }=\mathrm{\rho }\times \mathrm{\pi }\times {\left(28\times {10}^{-2}\right)}^2$

$\text{mass of removed disc }=\mathrm{\rho }\times \mathrm{\pi }\times {\left(21\times {10}^{-2}\right)}^2$

the center of the removed disc is a distance of 7cm from the center of the original disc.

${\mathrm{x}}_{\mathrm{com}}=\frac{\mathrm{\rho }\times \mathrm{\pi }\times {\left(28\times {10}^{-2}\right)}^2\times 0-\mathrm{\rho }\times \mathrm{\pi }\times {\left(21\times {10}^{-2}\right)}^2\times \left(7\times {10}^{-2}\right)}{\mathrm{\rho }\times \mathrm{\pi }\times {\left(28\times {10}^{-2}\right)}^2-\mathrm{\rho }\times \mathrm{\pi }\times {\left(21\times {10}^{-2}\right)}^2}=0.09$

${\mathrm{x}}_{\mathrm{com}}=9\mathrm{cm}$