Answer
$$ B_{net}=\dfrac{ 2\mu_0I}{ \pi l} \tag{Leftward}$$
Work Step by Step
According to the right-hand rule: the direction of the magnetic field exerted by the wire in the upper right corner (Point K in the figure below) at point C is directed toward the left upper wire (Point N in the figure below).
The direction of the magnetic field exerted by the wire in the upper left corner (N) at point C is directed toward the left lower wire (M).
The direction of the magnetic field exerted by the wire in the lower right corner (L) at point C is directed toward the left lower wire (M).
And the direction of the magnetic field exerted by the wire in the lower left corner (M) at point C is directed toward the left upper wire (N).
And since all the wires are carrying the same current and all at the same distances from point C, so the vertical components of these 4 fields cancel each other and the remaining components are the horizontal-western components only.
In the figure below, the dashed circle are not accurate but they make the imagination of the field direction easier.
Now we need to find the net electric field which is directed left (toward the west).
$$B_{net}=B_{Kx}+B_{Lx},B_{Mx}+B_{Nx}$$
$$B_{net}=\dfrac{\mu_0I}{2\pi r}\cos 45^\circ+\dfrac{\mu_0I}{2\pi r}\cos 45^\circ+\dfrac{\mu_0I}{2\pi r}\cos 45^\circ+\dfrac{\mu_0I}{2\pi r}\cos 45^\circ$$
$$B_{net}=4\left(\dfrac{\mu_0I}{2\pi r}\cos 45^\circ\right)\tag 1$$
whereas $r$ is the distance from any corner to the center point C which is given by
$$r=\sqrt{\left(\dfrac{l}{2}\right)^2+\left(\dfrac{l}{2}\right)^2}=\sqrt{\dfrac{2l^2}{4}}=\dfrac{\sqrt{2} }{2}l$$
Plugging into (1);
$$B_{net}=4\left(\dfrac{2\mu_0I}{2\sqrt2\pi l}\cos 45^\circ\right) $$
$$B_{net}= \dfrac{ 4\mu_0I}{ \sqrt2\pi l}\cdot \dfrac{\sqrt{2}}{2} $$
$$\boxed{B_{net}=\dfrac{ 2\mu_0I}{ \pi l}}\tag{Leftward}$$