Explanation

(d) Let $V$ be the volume of wooden ball. The mass of ball is $m = V\ \rho$
Upward acceleration,
$a = \frac{upward\ thrust - weight\ of\ ball}{mass\ of\ ball}$
$= \frac{V\ \rho_{0}g - V\rho g}{V\rho} = \frac{(\rho_{0} - \rho)g}{\rho}$
If $v$ is the velocity of ball on reaching the surface after being released at depth $h$ is
$v = \sqrt{2as} = \left\lbrack 2\left( \frac{\rho_{0} - \rho}{\rho} \right)gh \right\rbrack^{1/2}$
If $h'$ is the vertical distance reached by ball above the surface of water, then
$h^{'} = \frac{v^{2}}{2g} = \frac{2\left( \rho_{0} - \rho \right)}{\rho}gh \times \frac{1}{2g}\ $
$= \left( \frac{\rho_{0} - \rho}{\rho} \right)h = (\rho_{0}/\rho)$