(a)

Potential energy per unit volume $=\frac{1}{2} \times$ stress $\times$ strain

$=\frac{1}{2} \times \frac{F}{A} \times \frac{\Delta L}{L}$

So, Potential energy $=$ potential energy per unit volume $\times$ volume

$=\frac{1}{2} \times \frac{F \cdot \Delta L}{A \cdot L} \times A \cdot L$   $\{$ Volume $=$ Length $\times$ cross-sectional area $\}$

$\Delta U=\frac{1}{2} \cdot F \cdot \Delta L$  $\left\{\begin{array}{l}F=10 \times 10 \,N \\ \Delta L=10 \,mm =10 \times 10^{-3} \,m \end{array}\right.$

Substituting values

$\Delta U=\frac{1}{2} \times 100 \times \frac{10}{1000}$

$\Delta U=0.5 \,J$