Answer
See the detailed answer below.
Work Step by Step
We know, from Biot-Savart law, that the magnetic field of a moving charged particle is given by
$$B=\dfrac{\mu_0}{4\pi }\dfrac{qv\sin\theta}{r^2}$$
$$\color{blue}{\bf [a]}$$
At point $(1,0,0)$, $\theta=0^\circ$, so
$$B=\dfrac{\mu_0}{4\pi }\dfrac{qv(0)}{r^2}=\color{red}{\bf 0}\;\rm T$$
$$\color{blue}{\bf [b]}$$
At point $(0,1,0)$, $\theta=90^\circ$, so
$$B=\dfrac{(4\pi\times 10^{-7})}{4\pi }\dfrac{(1.6\times 10^{-19})(1\times 10^7)\sin90^\circ}{0.01^2} $$
$$B=\color{red}{\bf 1.6\times 10^{-15}}\;\rm T$$
and its direction is out of the page.
$$\color{blue}{\bf [c]}$$
At point $(0,-2,0)$, $\theta=90^\circ$, so
$$B=\dfrac{(4\pi\times 10^{-7})}{4\pi }\dfrac{(1.6\times 10^{-19})(1\times 10^7)\sin90^\circ}{0.02^2} $$
$$B=\color{red}{\bf 4\times 10^{-16}} \;\rm T$$
and its direction is into the page.
