A bob of mass m is suspended by a light string of length L . It is...

The velocity given is minimum, just enough to complete verticle circle. At the top most point, tension of the string will be zero and gravitational force will provide the required centripetal force.

Therefore, 

$mg = \dfrac{m{\left({V}_{H}\right)}^{2}}{L} ⇒ \dfrac{1}{2}m{\left({V}_{H}\right)}^{2} = \dfrac{1}{2}gL$

Apply energy conservation between point $A$ and $B$, we get

$\dfrac{1}{2}m{V}_{L}^{2} = \dfrac{1}{2}m{V}_{H}^{2} + mg ( 2L )$

$⇒ {V}_{L} = \sqrt{5gL}$

Also, ${V}_{H} = \sqrt{gL}$

Required ratio, $\dfrac{( K . E{)}_{A}}{( K . E{)}_{B}} = \dfrac{\dfrac{1}{2}m ( \sqrt{5gL}{)}^{2}}{\dfrac{1}{2}m ( \sqrt{gL}{)}^{2}} = \dfrac{5}{1}$