Restoring torque is

$\tau_1=F_b \sin \theta \times l=V\left(\rho_{ air }-\rho_{ He }\right) g l \sin \theta$

$=V\left(\rho_{ air }-\rho_{ He }\right) g l \theta$

For small angular displacements, $\sin \theta \approx \theta$

And the inertial torque an balloon is

$\tau_2=l \alpha=m l^2 \alpha=V \cdot \rho_{ He } \cdot l^2 \cdot \alpha$

Helium balloon can be viewed as a mass tied to end of a string.

Equating both torques, we have

$V \cdot \rho_{ He } \cdot l^2 \cdot \alpha=-V\left(\rho_{ air }-\rho_{ He }\right) l \theta$

(Negative sign appears as both torques are in opposite directions)

$\alpha=-\left(\frac{\rho_{ air }-\rho_{ He }}{\rho_{ He }}\right) \cdot \frac{g}{l} \cdot \theta$

So, if $\omega$ is angular frequency, then

$\omega^2=\left(\frac{\rho_{ air }-\rho_{ He }}{\rho_{ He }}\right) \cdot \frac{g}{l}$

$\therefore$ Time period of oscillations of balloon is

$T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\left(\frac{\rho_{ He }}{\rho_{ air }-\rho_{ He }}\right) \cdot \frac{l}{g}}$