Answer
$({{\bf -1.13\times 10^{-15}}\;\rm T})\;\hat k$
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, $\theta=45^\circ$, so
$$B=\dfrac{(4\pi\times 10^{-7})}{4\pi }\dfrac{(1.6\times 10^{-19})(2\times 10^7)\sin45^\circ}{(0.01^2+0.01^2)} $$
$$B=\color{red}{\bf 1.13\times 10^{-15}}\;\rm T$$
and its direction is into the page in the negative $z$-direction.
