Answer
b) is correct.
Work Step by Step
Forces $F_1$ and $F_4$ pass through the point where the axis of rotation also passes, so they produce no torques.
Take $l$ to be the distance from the axis of rotation to the point where forces $F_2$ and $F_3$ are applied and $\theta$ to be the angle $F_2$ makes with the door. Then we have $$\tau_2=Fl\sin\theta$$ $$\tau_3=Fl$$
Because $l\gt l\sin\theta$, $\tau_3\gt\tau_2$
So to rank the torques from largest to smallest, $\tau_3\gt\tau_2\gt\tau_1=\tau_4$
