A disc of mass 1   kg and radius R is free of rotate about a horizontal axis passing through its centre…

Let the angular speed of disc be $\omega$.

Using conservation of mechanical energy

$mg2R=\frac{1}{2}{I}_{\mathrm{disc} }{\omega }^2+\frac{1}{2}{I}_{\mathrm{particle} }{\omega }^2$

Where, $I$ is moment of inertia.

$\Rightarrow mg2R=\frac{{\omega }^2}{2}\left[\frac{m^2}{2}+m{R}^2\right]$

$\Rightarrow mg2R=\frac{{\omega }^2}{2}\left(\frac{3}{2}m{R}^2\right)$

$\Rightarrow \frac{3}{4}{\omega }^2=\frac{2g}{R}$             

 $\Rightarrow {\omega }^2=\frac{8g}{3R}$

Thus, angular speed is $\omega =\sqrt{\frac{80}{3R}} \mathrm{rad} {\mathrm{s}}^{-1}$

Given $\omega =4\sqrt{\frac{x}{3R}}$

Comparing both, $16\frac{x}{3R}=\frac{80}{3R}$

Therefore, the value of $x=5$.