The magnetic field at point $O$ due to the two semi-infinite portions of both wires lying along line $BE$ will be $0$.

Using right-hand rule, it can be determined that the magnetic field due to both the semi-infinite portions of wires $BC \text{and} ET$ which are perpendicular to $BE$ will be outwards from the plane.

The magnetic field due to one such portion will be:

$B=\frac{{\mu }_{0}I}{4\pi {\displaystyle \left(\frac{a}{2}\right)}}$

Since the magnetic field of both the portions is in same direction, the net magnetic field at $O$ will be:

$B_{net}=2\times \frac{{\mu }_{0}I}{4\pi {\displaystyle \left(\frac{a}{2}\right)}}=\frac{{\mu }_{0}I}{\pi a}$