The acceleration of the rod is

$a=\frac{F}{m}$

So, the tension in the rod at a distance $x$ from the free end is

$T\left(x\right)=\frac{mx}{l}\times \frac{F}{m}=\frac{Fx}{l}$

The longitudinal stress developed inside the rod at a distance $x$ from the free end is

$\sigma =\frac{T\left(x\right)}{A}=\frac{Fx}{lA}$

The strain energy density (energy per unit volume) is

$\frac{dU}{dV}=\frac{1}{2}\frac{{\sigma }^2}{Y}=\frac{1}{2}\frac{F^2{x}^2}{Y{A}^2{l}^2}$, where $dV=Adx$ is the volume of a small part of the rod

So the energy stored in the rod is

$U=\frac{1}{2}\frac{F^2}{Y{A}^2{l}^2}{\int }_0^l{x}^{2}dx$

$U=\frac{F^{2}l}{6A^{2}Y}$