Answer
$\vec B_{A}=7.8\mu T$ and it is directed out of the page. At point $B$, the magnetic field is zero. $\vec B_{C}=7.8\mu T$ and it is directed into the page.
Work Step by Step
We can calculate the magnetic field at the points A, B and C as follows:
$\vec B_A=\frac{\mu_{\circ}I}{2\pi}(\frac{1}{r_{1A}+\frac{1}{r_{2A}}})$
$\implies \vec B_A=\frac{(4\pi \times 10^{-7}T.m/A)(2.2A)}{2\pi}(\frac{1}{0.075m}+\frac{1}{0.225})$
$\vec B_A=7.8\mu T$ and it is directed out of the page.
We know that at point B, the magnetic field due to wire 1 is into the page and the magnetic field due to the wire 2 is out of the page. Thus, the net field at point B is zero.
The magnetic field at point C is given as
$\vec B_{C}=\frac{(4\pi\times 10^{-7}Tm/A )(2.2A)}{2\pi}(\frac{1}{0.225m}+\frac{1}{0.075m})$
$\vec B_{C}=7.8\mu T$ and it is directed into the page.
