Answer
$({{\bf 2.83\times 10^{-16}}\;\rm T})\;\hat k$
Work Step by Step
We know, from Biot-Savart law, that the magnetic field of a moving charged particle is given by
$$B=\dfrac{\mu_0}{4\pi }\dfrac{qv\sin\theta}{r^2}$$
Plug the known from the given graph,
At the given point, $\theta=90^\circ+45^\circ=135^\circ$, so
$$B=\dfrac{(4\pi\times 10^{-7})}{4\pi }\dfrac{(1.6\times 10^{-19})(2\times 10^7)\sin135^\circ}{2(0.01^2+0.01^2)} $$
$$B=\color{red}{\bf 2.83\times 10^{-16}}\;\rm T$$
and its direction is out of the page in the positive $z$-direction.
