A rod of mass m and length $\mathcal{l}$  is made to stand at an angle of  $60°$ with the vertical.  Potential energy Of the rod in this position is 

For any uniform rod, the mass is supposed to be concentrated at its centre.
$\therefore$   height of the mass from ground is, h=$(l/2)$ sin $30°$  
$\therefore$  Potential energy of the rod $=m\times g\times \frac{\mathcal{l}}{2}\sin 30°$ $=m\times g\times \frac{\mathcal{l}}{2}\times \frac{1}{2}=\frac{mg\mathcal{l}}{4}$