$\mathrm{E}=1 \times 10^{4}\, \mathrm{V} / \mathrm{m}^{2}$

Since the path of electron remains undeviated, $\mathrm{qvB}=\mathrm{q} \mathrm{E}$ or

$\mathrm{v}=\frac{\mathrm{E}}{\mathrm{B}}=\frac{1 \times 10^{14}}{2 \times 10^{-3}}=0.5 \times 10^{7}$

$=5 \times 10^{6}\, \mathrm{m} / \mathrm{s}$

If the electricfield is removed, the path of the charged particle is circular and magnetic field provides the necessary centripetal force.

i.e., $\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mathrm{Bev} \Rightarrow \quad \mathrm{r}=\frac{\mathrm{mv}}{\mathrm{Be}}$

$=\frac{9.1 \times 10^{-31} \times 5 \times 10^{6}}{2 \times 10^{-3} \times 1.6 \times 10^{-19}}$

$=14.3 \times 10^{-3}\, \mathrm{m}=1.43\, \mathrm{cm}$