Given ${\omega }_0=10\pi rad/s$

The acceleration for translation $a=\frac{f}{m}=\frac{\mu mg}{m}=\mu g$

$=0.2\times 9.8=1.96m/s^2$

Angular retardation, $\alpha =\frac{\tau }{I}=\frac{fR}{I}=\frac{\mu mg\times R}{I}$

For disc $I=\frac{m{R}^2}{2}$

$\alpha =\frac{\mu mgR}{{\displaystyle \frac{m{R}^2}{2}}}=\frac{2\mu g}{R}=\frac{2\times 0.2\times 9.8}{0.1}=39.2rad/s^2$

For ring $I=m{R}^2$

$\alpha =\frac{\mu mgR}{m{R}^2}=\frac{\mu g}{R}=1.96rad/s^2$

It $t$ is the time taken to start pure rolling, then

$$\begin{gathered} \omega ={\omega }_0-\alpha t, ...\left(1\right) \\ \nu =0+\alpha t, ...\left(2\right) \\ \nu =\omega R ...\left(3\right) \end{gathered}$$

By solving above equations we get

$t=\frac{{\omega }_0}{\left(\alpha +{\displaystyle \frac{a}{R}}\right)}$

Thus $t_{disc}=\frac{10\mathrm{\pi }}{\left(39.2+{\displaystyle \frac{1.96}{0.1}}\right)}=0.53s$

Similarly $t_{ring}=0.80s$

Obviously disc begins to roll earlier than the ring