#### Answer

c. To the left.

#### Work Step by Step

Recall that the electric force can be expressed as $ \displaystyle \vec{F}_E = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2} = q\vec{E}$
Because $\vec{F}_E$ is directly related to (proportional to) $\vec{E}$, the direction that the electric field is in is the same direction that the electric force is in if it exists (for positive charges that is).
In this question, the diagram shows us that there is an electron in an electric field pointing to the right. Thus, the electric force is $\vec{F}_E = q\vec{E} = {-e}\vec{E} = -e\vec{E}$
Since there is a negative sign, the electric force on the electron is in the opposite direction as the electric field (if we were dealing with a proton instead, then the electric force would be in the same direction as the electric field). The force is to the left.
Option c. is correct.