Velocity $\vec{v} = v\hat{i}$, Magnetic field $\vec{B} = B\hat{j}$.
For a positive charge: $\vec{F} = q(\vec{v} \times \vec{B}) = q(v\hat{i} \times B\hat{j}) = qvB\hat{k}$ (along $+z$).
But electron has charge $q = -e$:
$$\vec{F} = (-e)(v\hat{i} \times B\hat{j}) = -evB\hat{k}$$
Wait — $\hat{i} \times \hat{j} = \hat{k}$, so force on electron is along $-\hat{k}$? Re-checking:
$\vec{F} = q\vec{v} \times \vec{B} = (-e)(v\hat{i} \times B\hat{j}) = -evB(\hat{i}\times\hat{j}) = -evB\hat{k}$.
However the standard answer given is $+z$. Using $\hat{i} \times \hat{j} = \hat{k}$ and charge $= -e$: force is along $-\hat{k}$.
Correct answer is along $-z$ axis (option b).