Answer
Ranking from largest to smallest net torque: $0^o, 45^o, 90^o$
Work Step by Step
Take half the bar's length to be $L$. The net torque produced is $$\sum\tau=FL-FL\sin\phi=FL(1-\sin\phi)$$
For $\phi=90^o$, $\sin\phi=1$, so $\sum\tau=0$
For $\phi=45^o$, $\sin\phi=\frac{\sqrt2}{2}$, so $\sum\tau=\Big(1-\frac{\sqrt2}{2}\Big)FL$
For $\phi=0^o$, $\sin\phi=0$, so $\sum\tau=FL$
Ranking from largest to smallest net torque: $0^o, 45^o, 90^o$
