If the hydrometer were in equilibrium or floating, its weight will be balanced by the buoyancy force acting on it by the fluid. During its small oscillation, let us locate the hydrometer when it is a vertically downward distance $x$ from its equilibrium position.

Here, the net unbalanced force on the hydrometer is the excess buoyancy force directed upward.

Thus, $F_b=\rho Vg=\rho \pi r^{2}xg$.

Now, restoring force $ma=F_b$. So, restoring acceleration is $a=-\frac{\rho \pi r^{2}gx}{m}$.

For SHM, acceleration is $a=-{\omega }^{2}x$.

Then, we have, ${\omega }^2=\frac{\rho \pi r^{2}g}{m}$.

Thus, the time period of oscillation is $T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{\pi r^2\rho g}}$.