A solid cylinder of radius R rolls without slipping with a center...

Let the mass of the cylinder be $m$ and its radius be $R$.

When the cylinder is rolling on the horizontal surface, its kinetic energy is:
$K_i = \dfrac{1}{2}mv_0^2 + \dfrac{1}{2}I_{cm}\omega_0^2$

Since it rolls without slipping, $\omega_0 = \dfrac{v_0}{R}$ and $I_{cm} = \dfrac{1}{2}mR^2$.
$K_i = \dfrac{1}{2}mv_0^2 + \dfrac{1}{2}\left(\dfrac{1}{2}mR^2\right)\left(\dfrac{v_0}{R}\right)^2 = \dfrac{3}{4}mv_0^2$

Given $v_0 = \sqrt{\dfrac{gR}{3}}$, we have $v_0^2 = \dfrac{gR}{3}$.
$K_i = \dfrac{3}{4}m\left(\dfrac{gR}{3}\right) = \dfrac{1}{4}mgR$

Let the corner be the reference level for potential energy. The initial total mechanical energy of the cylinder just as it reaches the corner is:
$E_i = K_i + U_i = \dfrac{1}{4}mgR + mgR = \dfrac{5}{4}mgR$

As the cylinder rotates around the corner, let $\theta$ be the angle the line connecting the corner to the center of mass makes with the vertical. Let $v$ be the speed of the center of mass at this angle.
The total mechanical energy at angle $\theta$ is:
$E_f = \dfrac{3}{4}mv^2 + mgR\cos\theta$

By conservation of mechanical energy, $E_i = E_f$:
$\dfrac{5}{4}mgR = \dfrac{3}{4}mv^2 + mgR\cos\theta$

The forces acting on the cylinder along the radial direction (towards the corner) are the component of gravity $mg\cos\theta$ and the normal force $N$. The equation of motion is:
$mg\cos\theta - N = \dfrac{mv^2}{R}$

The cylinder loses contact with the corner when the normal force becomes zero ($N = 0$):
$mg\cos\theta = \dfrac{mv^2}{R} \Rightarrow mgR\cos\theta = mv^2$

Substituting this into the energy conservation equation:
$\dfrac{5}{4}mgR = \dfrac{3}{4}mv^2 + mv^2$

$\dfrac{5}{4}mgR = \dfrac{7}{4}mv^2$

Solving for $v$:
$v^2 = \dfrac{5}{7}gR$

$v = \sqrt{\dfrac{5gR}{7}}$

Answer: $\sqrt{\dfrac{5gR}{7}}$