Answer
b) is correct.
Work Step by Step
The work done by a force $F$ on the box is $$W=(F\cos\theta)s$$
As we can see here, $F_1$ acts directly on the box's rightward motion, so $\cos\theta=\cos0=1$, therefore, $W_1=F_1s$, while for $F_2$, $W_2=(F_2\cos\theta)s$
It can also be easily seen that $F_1\gt F_2\cos\theta$, because the vector is longer. Therefore, $W_1\gt W_2$, and b) is correct.
