Using the conservation of mechanical energy between $\mathrm{A}$ and $\mathrm{B}$

At B, total kinetic energy = $mgh$

Here,  $m$ = mass of the ball

The ratio of rotational to translational kinetic energy would be,

        $\frac{K_R}{K_T}=\frac{2}{5}$

∴     $K{}_R=\frac{2}{7}\mathit{\text{mg}}h\text{and}K{}_T=\frac{5}{7}\mathit{\text{mgh}}$

In portion $\mathrm{BC}$, friction is absent. Therefore, rotational kinetic energy will remain constant and translational kinetic energy will convert into potential energy.

Hence, if $H$ be the height to which ball climbs in $\mathrm{BC}$, then

     $mgH = K_{\mathrm{T}}$

or    $\mathit{\text{mgH}}=\frac{5}{7}\mathit{\text{mg}}h\text{or}H=\frac{5}{7}h$