Let normal reaction on the rod due to boy is $N_1$ and due to man is $N_2$
Given, $N_1=\frac{W}{4}$
$\therefore$ $N_2=W-\frac{W}{4}=\frac{3W}{4}$



Let $x$ be the distance of the man from the other end.

For rotational equilibrium net torque must be zero.
$\Rightarrow N_1\times \frac{L}{2}=N_2\times \left(\frac{L}{2}-x\right)$
$\Rightarrow$ $\frac{W}{4}\times \frac{L}{2}=\frac{3W}{4}\times \left(\frac{L}{2}-x\right)$
$\Rightarrow$ $x=\frac{L}{3}$