Work done in stretching a spring,

$\mathrm{W}=\frac{1}{2} \mathrm{~F} \times \Delta l$

where $\Delta l$ increases in length due to applying force,

$\therefore \mathrm{W} \propto \Delta l \quad\left[\because \frac{1}{2}, \mathrm{~F} \text { constant }\right]$

$\therefore \mathrm{W} \propto \frac{1}{\mathrm{Y}}\left[\because \Delta l=\frac{\mathrm{F} l}{\mathrm{AY}}, \mathrm{F}, l, \text { A are constant }\right]$

$\therefore \frac{\mathrm{W}_{\text {steel }}}{\mathrm{W}_{\text {copper }}}=\frac{\mathrm{Y}_{\text {copper }}}{\mathrm{Y}_{\text {steel }}}$

$\text { but } \mathrm{Y}_{\text {steel }}>\mathrm{Y}_{\text {copper }}$

$\therefore \mathrm{W}_{\text {steel }}<\mathrm{W}_{\text {copper }}$

$\therefore$ Hence, more work will be done in case of spring made of copper.