#### Answer

$B = (0,0,-0.11~T)$

#### Work Step by Step

We can find the magnetic field at the point:
$B = \frac{\mu_0~q~v~sin~\theta}{4~\pi~r^2}$
$B = \frac{(4\pi\times 10^7~H/m)~(1.6\times 10^{-19}~C)~(2.0\times 10^7~m/s)~sin~45^{\circ}}{(4~\pi)~(\sqrt{2}\times 0.010~m)^2}$
$B = 0.11~T$
By the right hand rule, the magnetic field is in the $-z$ direction.
We can write the magnetic field as a vector:
$B = (0,0,-0.11~T)$