Answer
$ \tau_f\gt \tau_a=\tau_b\gt \tau_c=\tau_d\gt \tau_f $
Work Step by Step
We know that torque is given by
$$\tau=F_{\perp }r$$
where $F_{\perp }$ is the perpendicular force component to $r$.
Now we need to find the torque produced by each force around the hang point of the door.
$$\tau_a=F\left[\dfrac{L}{2}\right]=\dfrac{FL}{2}$$
$$\tau_b=2F\left[\dfrac{L}{4}\right]=\dfrac{FL}{2}$$
$$\tau_c=F\sin45^\circ\left[\dfrac{L}{2}\right]=F\dfrac{\sqrt2}{2}\dfrac{ L}{2}=\dfrac{\sqrt2 FL}{4}=0.35FL$$
$$\tau_d=F\sin45^\circ\left[\dfrac{L}{2}\right]=F\dfrac{\sqrt2}{2}\dfrac{ L}{2}=\dfrac{\sqrt2 FL}{4}=0.35FL$$
$$\tau_e=2F\left[L\right]=2FL$$
$$\tau_f= F\left[L\right]\sin 0^\circ=0$$
Thus, from above, the rank of the torques is
$$\boxed{\tau_f\gt \tau_a=\tau_b\gt \tau_c=\tau_d\gt \tau_f}$$
