Let mass per unit area of the disc $=m$
Total mass of disc, $M=\pi R^2 m$
Mass of the scooped out portion of the disc,
$M^{\prime}=\pi r^2 m$
For small objects, centre of gravity and centre of mass are at same position.
We take the centre $O$ of the disc as the origin. The masses $M$ and $M^{\prime}$ can be supposed to be concentrated at the centres of the disc and scooped out portion respectively. The mass $M^{\prime}$ of the removed portion is taken negative.


The $x$-coordinate of the centre of mass of the remaining portion of the disc will be
$\begin{aligned}
x_{\mathrm{CM}} & =\frac{M x_1-M^{\prime} x_2}{M-M^{\prime}}=\frac{M \times 0-M^{\prime} \times 3}{\pi R^2 m-\pi r^2 m}=\frac{-3 M^{\prime}}{\pi m\left(R^2-r^2\right)} \\
& =\frac{-3 \pi r^2 m}{\pi m\left(R^2-r^2\right)}=-\frac{-3 r^2}{R^2-r^2}=\frac{-3 \times 3^2}{6^2-3^2}=-1 \mathrm{~cm} \\
\Rightarrow x_{\mathrm{CM}} & =-\mathrm{lcm}
\end{aligned}$
Hence, distance of centre of gravity of the resulting flat body from the centre of the original disc will be $1 \mathrm{~cm}$ left side.