Answer
The net torque about the axle is 5.0 N m
Work Step by Step
We can find the net torque about the axle. Let $F_1$ be the 50-N force which is directed downward.
$\tau_{net} = \sum \tau$
$\tau_{net} = r_1~F_1 + r_2~F_2$
$\tau_{net} = (0)(50~N)+(0.10~m)(50~N)$
$\tau_{net} = 5.0~N~m$
The net torque about the axle is 5.0 N m.
