DThe same stress but different strains
d
(d) ${\rm{Stress}} = \frac{{{\rm{Force}}}}{{{\rm{area}}}}$.
In the present case, force applied and area of cross-section of wires are same, therefore stress has to be the same.
${\rm{Strain}} = \frac{{{\rm{Stress}}}}{Y}$
Since the Young’s modulus of steel wire is greater than the copper wire, therefore, strain in case of steel wire is less than that in case of copper wire.