A wire with a total length L is used to form a square loop carrying...

Let the side length of the square loop be $a$.

Since the total length of the wire is $L$, the perimeter of the square is $4a = L \Rightarrow a = \dfrac{L}{4}$.

The area of the square loop is $A = a^2 = \left(\dfrac{L}{4}\right)^2 = \dfrac{L^2}{16}$.

The magnetic dipole moment of the loop is $M = iA = \dfrac{iL^2}{16}$.

The magnetic field $B$ is parallel to the plane of the loop, so the angle between the area vector of the loop and the magnetic field is $\theta = 90^{{}^{{}^{\circ}}}$.

The magnitude of the torque on the loop is $\tau = MB\sin\theta$.

Substituting the values, we get $\tau = \left(\dfrac{iL^2}{16}\right)B\sin 90^{{}^{{}^{\circ}}} = \dfrac{iL^2B}{16}$.