Condition for losing contact. While pivoting about the corner, contact is lost when the required centripetal force exceeds the available radial component of gravity: $m{\omega }^{2}R=mg\cos \theta$, i.e. ${\omega }^{2}R=g\cos \theta$ at the moment of separation, where $\theta$ is the angle of the line from corner to centre measured from vertical.

Initial angle. Initially, ${\omega }_0=v_0/R=\sqrt{g/(3R)}$, and the cylinder pivots with the contact line vertical, so ${\theta }_0=0$. The condition ${\omega }_0^{2}R=g\cos {\theta }_0$ would require $g/3=g$, which is not satisfied, so initially the cylinder stays in contact and rotates.

Energy conservation about the pivot. Loss of height of the centre as it swings through angle $\theta$ is $R(1-\cos \theta )$. With moment of inertia about the corner $I=\frac{1}{2}m{R}^2+m{R}^2=\frac{3}{2}m{R}^2$:

$mgR(1-\cos \theta )=\frac{1}{2}\cdot \frac{3}{2}m{R}^2({\omega }^2-{\omega }_0^2)$.

Substitute the separation condition ${\omega }^{2}R=g\cos \theta$ and ${\omega }_0^2=g/(3R)$:

$g-g\cos \theta =\frac{3}{4}R{\omega }^2-\frac{g}{4}\Rightarrow g-{\omega }^{2}R=\frac{3}{4}R{\omega }^2-\frac{g}{4}$.

$\frac{5g}{4}=\frac{7}{4}R{\omega }^2\Rightarrow {\omega }^2=\frac{5g}{7R}$.

Centre-of-mass speed at separation: $v=R\omega =\sqrt{\frac{5gR}{7}}$. Option (B).