Elastic potential energy V is stored in the wire when a force F is applied and the extension $\Delta l$
$V= \large\frac{1}{2}$$ F \Delta l$
$\quad= \large\frac{1}{2} \frac{F}{A}$$ \times A \times \large\frac{\Delta l}{l}$$ \times l$
$\quad=\large\frac {1}{2}$ stress $\times$ strain $\times$ volume $(A \times l=volume)$
$\quad=\large\frac{1}{2}$$ y \;strain ^2 \times volume$ $\qquad[Y=\large\frac{stress}{strain}]$
Hence d is the corrwect answer.