(A) The Young's modulus $(Y)$ is defined as the ratio of stress to strain,given by $Y = \frac{F \cdot l}{A \cdot \Delta l}$.
Consider a steel wire and a plastic wire of the same length $(l)$,cross-sectional area $(A)$,subjected to the same deforming force $(F)$.
For steel: $Y_{S} = \frac{F \cdot l}{A \cdot \Delta l_{S}}$
For plastic: $Y_{P} = \frac{F \cdot l}{A \cdot \Delta l_{P}}$
Since plastic undergoes a larger extension than steel for the same force,$\Delta l_{P} > \Delta l_{S}$.
Taking the ratio: $\frac{Y_{S}}{Y_{P}} = \frac{\Delta l_{P}}{\Delta l_{S}}$.
Since $\Delta l_{P} > \Delta l_{S}$,it follows that $\frac{Y_{S}}{Y_{P}} > 1$,which implies $Y_{S} > Y_{P}$.
Elasticity is defined by the ability of a material to resist deformation,which is directly proportional to Young's modulus. Therefore,steel is more elastic than plastic.