Answer
$({\bf 6.25\times 10^6}{\;\rm m/s})\;\hat z$
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 for the first point,
$$1\times 10^{-13}\;\hat j=\dfrac{(4\pi\times 10^{-7})}{4\pi }\dfrac{(1.6\times 10^{-19})v\sin\theta_1}{0.001^2} $$
This one the magnetic field is in the positive $y$-direction which means that the two vectors of $\vec v$ and $\vec r$ are in the $x$-$z$ plane.
Hence, the velocity vector is in the positive $z$-direction.
Plug the known for the second point,
$$-1\times 10^{-13}\;\hat i=\dfrac{(4\pi\times 10^{-7})}{4\pi }\dfrac{(1.6\times 10^{-19})v\sin\theta_2}{0.001^2} $$
Here the magnetic field is in the negative $x$-direction which means that the two vectors of $\vec v$ and $\vec r$ are in the $y$-$z$ plane.
Hence, the velocity vector is in the positive $z$-direction as well.
From all the above, $\theta_1=\theta_2=90^\circ$, and hence the velocity is then
$$1\times 10^{-13} =\dfrac{(4\pi\times 10^{-7})}{4\pi }\dfrac{(1.6\times 10^{-19})v\sin90^\circ}{0.001^2} $$
$$v=\dfrac{1\times 10^{-13}\times 0.001^2}{10^{-7}(1.6\times 10^{-19})}$$
$$v=(\color{red}{\bf 6.25\times 10^6}{\;\rm m/s})\;\hat z$$
