A square coil with n turns and an edge length l carries a current i...

The magnetic moment of the square coil is given by $\mu = n i A = n i l^2$.

Since the magnetic field $B$ is parallel to one of the edges of the coil, it lies in the plane of the coil. The angle between the magnetic moment vector $\vec{\mu}$ (which is perpendicular to the plane of the coil) and the magnetic field $\vec{B}$ is $90^{\circ}$.

The torque exerted by the magnetic field on the coil is $\tau_{B} = \mu B \sin 90^{\circ} = n i l^2 B$.

This magnetic torque tends to lift one edge of the coil and tip it over the opposite edge. The restoring torque is provided by the weight of the coil.

The weight $M g$ acts through the center of mass of the square coil, which is at a perpendicular distance of $\dfrac{l}{2}$ from any edge.

The restoring torque about the tipping edge is $\tau_{g} = M g$\left(\dfrac{l}{2}\right).

For the coil to just start tipping over, the magnetic torque must be equal to or greater than the restoring torque:

$\tau_{B} \ge \tau_{g}$

$n i l^2 B \ge$\dfrac{M g l}{2}

$B \ge$\dfrac{M g}{2 n i l}

Thus, the minimum value of the magnetic field required is \dfrac{M g}{2 n i l}.