Weight of suspended mass:

$W_1=m_1\,g$

only half of the weight of the rod acting at the mid point,

$W _2=\frac{ m _2 g }{2}$

strees at the mid point :

$\text { Stress } =\frac{ W _1+ W _2}{ A }$

$\Rightarrow \frac{ m _1 g +\frac{ m _2 g }{2}}{ A }$

$\Rightarrow \frac{\frac{2 m _1 g + m _2 g }{2}}{ A }$

$\therefore \frac{ g \left(2 m _1+ m _2\right)}{2 A }$