An electric current i is carried by a wire of length l which is...

Let the side length of the square be $a$. Since the total length of the wire is $l$, we have $4a = l \Rightarrow a = \dfrac{l}{4}$.

The magnetic field at the centre of the square due to one of its sides is given by the formula for a finite straight wire:

$B_1 = \dfrac{\mu_0 i}{4\pi d} (\sin\theta_1 + \sin\theta_2)$

The perpendicular distance from the centre to the side is $d = \dfrac{a}{2} = \dfrac{l}{8}$.

The angles subtended by the ends of the side at the centre are \theta_1 = \theta_2 = 45^{}^{{}^{\circ}}.

Substituting these values into the formula:

B_1 = \dfrac{\mu_0 i}{4\pi \left(\dfrac{l}{8}\right)} (\sin 45^{}^{{}^{\circ}} + \sin 45^{}^{{}^{\circ}})

$B_1 = \dfrac{2\mu_0 i}{\pi l} \left(\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}}\right) = \dfrac{2\mu_0 i}{\pi l} \left(\dfrac{2}{\sqrt{2}}\right) = \dfrac{2\sqrt{2}\mu_0 i}{\pi l}$

Since the square has four identical sides and the magnetic field due to each side points in the same direction at the centre, the total magnetic field is:

$B = 4B_1 = 4 \times \dfrac{2\sqrt{2}\mu_0 i}{\pi l} = \dfrac{8\sqrt{2}\mu_0 i}{\pi l}$