Solution

Let's consider a small deviation from equilibrium by an angle $\alpha$:

The moment acting on the frame from the magnetic field: $$M = BIS\cdot \sin\alpha = BIa^2 \sin\alpha$$ where $a$ is the side of the frame

gravity does not create a moment as it passes through the point of rotation

From the basic equation of rotational motion: $$J\ddot{\alpha} = \sum\pm{M_i}$$ for our case: $$J\ddot{\alpha} = -BIa^2 \sin\alpha\quad (1)$$ now we will find the moment of inertia $J$ of the frame during the observed oscillations

The moment of inertia of the upper and lower sides of the frame is $$J_1=J_2 = \frac{m}{4}\cdot \left(\frac{a}{2}\right)^2 = \frac{m a^2}{16}\quad (2)$$ where $a/2$ is the distance to this side.

Moment of inertia of the lateral sides: $$J_3 = J_4 = \frac{1}{12} \cdot \frac{m}{4}\cdot a^2 = \frac{m a^2}{48}\quad (3)$$ We find the moment of inertia $J$ by substituting $(2)$ and $(3)$ $$J = J_1 + J_2 + J_3 + J_4$$ $$J = 2\cdot \frac{m a^2}{16} + 2 \cdot \frac{m a^2}{48} = \frac{m a^2}{6}$$ then equation $(1)$ will look like: $$\frac{m a^2}{6}\ddot{\alpha} = -BIa^2 \sin\alpha$$ $$\ddot{\alpha} = -\frac{6BI}{m} \alpha (t)$$ Solving the equation of harmonic oscillations, we find the oscillation frequency $$\boxed{\omega = \sqrt{\frac{6BI}{m}}}$$