According to the question,
$\therefore $ Kinetic energy for the rolling disc,
or $KE _{r}=\frac{1}{2} m v_{1}^{2}+\frac{1}{2} I \omega^{2}$
$\left(\therefore \right.$ Moment of inertia, $\left.I=\frac{m R^{2}}{2}\right)$
or $=\frac{1}{2} m v_{1}^{2}+\frac{1}{2} \frac{m R^{2}}{2}\left(\frac{v_{1}}{R}\right)^{2}$
Or $KE _{r}=\frac{3}{4} m v_{1}^{2}\,\,\,...(i)$
Now, $KE$ for the sliding disc,
$\therefore KE _{s}=\frac{1}{2} m v_{2}^{2}\,\,\,...(ii)$
Given, $KE$ of rolling disc$= KE$ of sliding disc
Or, $\frac{3}{4} m v_{1}^{2}=\frac{1}{2} m v_{2}^{2}$ or $\frac{v_{1}^{2}}{v_{2}^{2}}=\frac{2}{3}$
or $\frac{v_{1}}{v_{2}}=\sqrt{\frac{2}{3}}$
or $v_{1}: v_{2}=\sqrt{2}: \sqrt{3}$