Answer
$B_{net} = (8.0\times 10^{-5}~T)~\hat{j}$
Work Step by Step
The distance from each current to the center is $\frac{a}{\sqrt{2}}$
By the right hand rule and symmetry, the horizontal components of the magnetic fields due to the currents cancel out, and the net magnetic field at the center is the sum of the vertical components which are all directed upward.
We can find the vertical component of the magnetic field at the center due to current in the top left corner:
$B = \frac{\mu_0~i}{2~\pi~R}~cos~45^{\circ}$
$B = \frac{(4\pi\times 10^{-7}~H/m)(20~A)}{(2~\pi)~(0.20~m/\sqrt{2})}~\frac{1}{\sqrt{2}}$
$B = \frac{(4\pi\times 10^{-7}~H/m)(20~A)}{(2~\pi)~(0.20~m)}$
$B = 2.0\times 10^{-5}~T$
By symmetry, the vertical component of the magnetic field at the center due to each of the other currents is also this value.
We can find the net magnetic field at the center:
$B_{net} = (4)(2.0\times 10^{-5}~T)$
$B_{net} = 8.0\times 10^{-5}~T$
Note that the net magnetic field is directed upward. We can express the net magnetic field at the center in unit-vector notation:
$B_{net} = (8.0\times 10^{-5}~T)~\hat{j}$
