Since from the given figure, in order to complete the vertical circle, velocity at the bottom of vertical circle should be $\sqrt {5gR}$.

As body is at rest initially, i.e., speed = 0.
At point A, speed = v.
As track is frictionless, so total mechanical energy will remain constant.

$\therefore$ 0 + mgh = ${1 \over 2}m{v^2}$ + 0

$\Rightarrow$ h = ${{{v^2}} \over {2g}}$

For completing the vertical circle, v $\ge$ $\sqrt {5gR}$

$\therefore$ h = ${{{5gR} \over {2g}}}$ = ${5 \over 2}R$ = ${5 \over 4}D$