Answer
$B = 5.0\times 10^{-6}~T$
Work Step by Step
We can write the expression for the magnetic field produced by a current in a straight wire:
$B = \frac{\mu_0~i}{2~\pi~R}$
We can find the magnetic field at the point $(0, 2.0~m, 0)$ due to the current in the +x direction:
$B = \frac{\mu_0~i}{2~\pi~R}$
$B = \frac{(4\pi\times 10^{-7}~H/m)(30~A)}{(2~\pi)~(2.0~m)}$
$B = 3.0\times 10^{-6}~T$
By the right hand rule, this magnetic field is in the +z direction.
We can find the magnetic field at the point $(0, 2.0~m, 0)$ due to the current in the +z direction:
$B = \frac{\mu_0~i}{2~\pi~R}$
$B = \frac{(4\pi\times 10^{-7}~H/m)(40~A)}{(2~\pi)~(2.0~m)}$
$B = 4.0\times 10^{-6}~T$
By the right hand rule, this magnetic field is in the -x direction.
We can find the net magnetic field at the point $(0, 2.0~m, 0)$:
$B = \sqrt{(3.0\times 10^{-6}~T)^2+(4.0\times 10^{-6}~T)^2}$
$B = 5.0\times 10^{-6}~T$
