Answer
$\tau_{max}=4.25N.m$
Work Step by Step
We have $$\tau=Fl$$
Because $F$ is constant here, the change in torque depends on the change in the lever arm.
In a square, the diagonal is the longest line. So the lever arm $l$ is maximum when the lever arm is half this diagonal. In other words, $$l_{max}=\frac{1}{2}\sqrt{0.4^2+0.4^2}=0.283m$$
Therefore, $$\tau_{max}=(15N)\times(0.283m)=4.25N.m$$
