Explanation

(a) As a is the side of cube $\sigma$ is its density
Mass of cube $a^{3}\sigma,$ its weight $= a^{3}\sigma g$
Let $h$ be the height of cube immersed in liquid of density of $\rho$ in equilibrium then, $F = a^{2}h\rho g = Mg = a^{3}\sigma g$
If it is pushed down by $y$ then the buoyant force
$F^{'} = a^{2}(h + y)\rho g$
Restoring force is $\mathrm{\Delta}F = F^{'} - F = a^{2}(h + y)\rho g - a^{2}h\rho g$
$= a^{2}y\ \rho\ g$
Restoring acceleration $= \frac{\mathrm{\Delta}F}{M} = - \frac{a^{2}y\rho g}{M} = - \frac{a^{2}\rho g}{a^{3}\rho}y$
Motion is S.H.M.
$\Rightarrow T = 2\pi\sqrt{\frac{a^{3}\sigma}{a^{2}\rho g}} = 2\pi\sqrt{\frac{a\sigma}{\rho g}}$