Answer
$B = (0,0,0.28\times 10^{-5})~T$
Work Step by Step
We can find the magnetic field at the point:
$B = \frac{\mu_0~q~v~sin~\theta}{4~\pi~r^2}$
$B = \frac{(4\pi\times 10^7~H/m)~(1.6\times 10^{-19}~C)~(2.0\times 10^7~m/s)~sin~45^{\circ}}{(4~\pi)~(2~\sqrt{2}~m)^2}$
$B = 0.28\times 10^{-5}~T$
By the right hand rule, the magnetic field is in the $+z$ direction.
We can write the magnetic field as a vector:
$B = (0,0,0.28\times 10^{-5})~T$
