Answer
$\tau_{net} = -0.94~N~m$
Work Step by Step
We can find the net torque about the axle as:
$\tau_{net} = \sum \tau$
$\tau_{net} = (r_1\times F_1) + (r_2\times F_2)+(r_3\times F_3)+(r_4\times F_4)$
$\tau_{net} = -(0.10~m)(30~N)+(0.05~m)(30~N)~sin(45^{\circ})+(0.05~m)(20~N)+(0.05~m)(20~N)~sin(0^{\circ})$
$\tau_{net} = -0.94~N~m$
