The moment of inertia of the system about instantaneous axis of rotation is given by,

$\mathrm{I}={\mathrm{I}}_{\mathrm{ring}}+{\mathrm{I}}_{\mathrm{rod}}$

$\Rightarrow \mathrm{I}=({\mathrm{MR}}^2+{\mathrm{MR}}^2)+\left(\frac{1}{12}{\mathrm{MR}}^2+{\mathrm{MR}}_1^2\right)$       (From parallel axis theorem)

$\Rightarrow \mathrm{I}=2{\mathrm{MR}}^2+\frac{1}{12}{\mathrm{MR}}^2+\mathrm{M}{\left(\frac{\sqrt{5}\mathrm{R}}{2}\right)}^2$           $\left(\because {\mathrm{R}}_1=\sqrt{{\mathrm{R}}^2+{\left(\frac{\mathrm{R}}{2}\right)}^2}=\frac{\sqrt{5}\mathrm{R}}{2}\right)$

$\Rightarrow \mathrm{I}=2{\mathrm{MR}}^2+\frac{1}{12}{\mathrm{MR}}^2+\frac{5}{4}{\mathrm{MR}}^2=2{\mathrm{MR}}^2+\frac{4}{3}{\mathrm{MR}}^2=\frac{10}{3}{\mathrm{MR}}^2$

Hence, ${\mathrm{I}}_{\mathrm{system}}=\frac{10}{3}{\mathrm{MR}}^2$.


Now from work - energy theorem, we get,

$\mathrm{Mg}\frac{\mathrm{R}}{2}=$Rotational kinetic energy of the system about instantaneous axis of rotation
(here $\frac{\mathrm{R}}{2}$ is the distance through which the center of mass of the rod descends)

$\Rightarrow \mathrm{Mg}\frac{\mathrm{R}}{2}=\frac{1}{2}\mathrm{ }{\mathrm{I}}_{\mathrm{system}}\mathrm{ }{\mathrm{\omega }}^2$

$\Rightarrow \mathrm{Mg}\frac{\mathrm{R}}{2}=\frac{1}{2}\times \frac{10}{3}{\mathrm{MR}}^2\times {\mathrm{\omega }}^2$

$\Rightarrow \mathrm{\omega }=\sqrt{\frac{3\mathrm{g}}{10\mathrm{R}}}$


Now velocity of center of mass is given as, $\mathrm{v}=\mathrm{R\omega }$.
$\Rightarrow \mathrm{v}=\mathrm{R}\times \sqrt{\frac{3\mathrm{g}}{10\mathrm{R}}}$
$\Rightarrow \mathrm{v}=\sqrt{\frac{3\mathrm{gR}}{10}}$