#### Answer

The electron is moving horizontally at an angle of $20.9^{\circ}$ east of south or $20.9^{\circ}$ west of north.

#### Work Step by Step

We can find the angle $\theta$ between the velocity vector and the magnetic field:
$F = q~v ~B~sin~\theta$
$sin~\theta = \frac{F}{q~v~B}$
$\theta = arcsin~(\frac{F}{q~v~B})$
$\theta = arcsin~\left(\frac{1.6 \times 10^{-14}~N}{(1.6\times 10^{-19}~C)(2.0\times 10^5~m/s)(1.4~T)}~\right)$
$\theta = 20.9^{\circ}$
Since the charge of the electron is negative, by the right hand rule, the velocity vector must be directed horizontally at an angle of $20.9^{\circ}$ east of south or $20.9^{\circ}$ west of north.