The moment of inertia  of a thin disc of mass $m$ and radius $R$ with a hole of radius $R/2$ about the Z-axis (passing through the center of hole), if $m=6kg,R=2m.$, is $\left(20+\mathrm{x}\right){\mathrm{kgm}}^2$. The value of $\mathrm{x}$ is

The easiest way is to “fill” the hole, and then subtract the contribution of the hole. 

The area of the full disc is $\pi R^2.$ 

The area of the hole is $\frac{\pi R^2}{4}.$ 

The area of the disc with the hole is $\frac{3\pi R^2}{4}.$ 

Thus by filling the hole, one increases the mass by a factor: 

$\frac{\pi R^2}{3\pi R^2/4}=\frac{4}{3}.$ 

The moment of inertia of a full disc with a mass of $\frac{4}{3}m$ around the z-axis is derived from the parallel axis theorem, by: $I_{\text{ƒull}}=\frac{1}{2}.\frac{4}{3}m{R}^2+\frac{4}{3}m{\left(\frac{R}{2}\right)}^2=m{R}^2\mathrm{.......}\left(i\right)$

Similarly, the mass of the small disc which fills the hole is given by: 

$m_1=\frac{\frac{\pi m{R}^2}{4}}{\frac{3\pi R^2}{4}}=\frac{1}{3}m\mathrm{.....}\left(ii\right),$

 Its moment of inertia around the axis is 

$I_{hole}=\frac{1}{2}.\frac{1}{3}m{\left(\frac{R}{2}\right)}^2=\frac{1}{24}m{R}^2\text{\hspace{0.17em}}\mathrm{......}\left(iii\right)$

Thus, we obtain: 

 $I_{total}=I_{\text{ƒull}}-I_{hole}=\frac{23}{24}m{R}^2\text{\hspace{0.17em}}\mathrm{......}\left(iv\right)$