Let the tension at point $A$ be $T_{A} .$ So, from Newton's second law 

$T_{A}-m g=\frac{m v_{c}^{2}}{R}$
Energy at point $A=\frac{1}{2} m v_{0}^{2}$   $…(i)$

Energy at point $C$ is

$\frac{1}{2} m v_{c}^{2}+m g \times 2 R \ldots .$         $…(ii)$
Applying Newton's second law at point $C$

$T_{c}+m g=\frac{m v_{c}^{2}}{R}$
To complete the loop $T_{c} \geq 0$

So, $m g=\frac{m v_{c}^{2}}{R}$

$\Rightarrow \quad v_{c}=\sqrt{g R}$      $…(ii)$

From Eqs. $(i)$ and $(ii)$ by conservation of energy 

$\frac{1}{2} m v_{0}^{2}=\frac{1}{2} m v_{c}^{2}+2 m g R$

$\Rightarrow \quad \frac{1}{2} m v_{0}^{2}=\frac{1}{2} m g R+2 m g R \quad\left(\because v_{c}=\sqrt{g R}\right)$

$\Rightarrow \quad v_{0}^{2}=g R+4 g R$

$\Rightarrow \quad v_{0}=\sqrt{5 g R}$