For practical reasons, we can consider each side of the square as an infinite wire. This can be seen on the following drawing:
This way, the field on the center will be the sum of the contribution of each wire. We can calculate the contribution of a single wire as:
[tex]B=\frac{\mu_0i}{2\pi d}=\frac{4\pi *10^{-7}*3.684}{2\pi(\frac{16.166*10^{-2}}{2})}=9.115*10^{-6}T[/tex]Then, the total field will be this, multiplied by the number of wires:
[tex]B_t=4*9.115*10^{-6}=36.46\mu T[/tex]Then, the resulting field will be Bt=36.46uT
