Two wires $A$ and $B$ of same length are made of same material. The figure represents the load $F$ versus extension $∆x$ graph for the two wires. Then, 

$$\begin{gathered} ∆x=\frac{Fl}{AY} \\ \therefore F=\left(\frac{AY}{L}\right)∆x \end{gathered}$$

i.e. $F$ versus $∆x$ graph is a straight line of slope $\frac{YA}{L}$.

                     (Slope)_B   >  (Slope)_A

$\therefore$                   ${\left(\frac{YA}{L}\right)}_B>{\left(\frac{YA}{L}\right)}_A$   or   ${\left(A\right)}_B>{\left(A\right)}_A$

They are of same material.

Hence,                 $Y_B=Y_A$

For a given load , strain produced in A  (${\in }_A)$is greater than that in B (${\in }_B)$. 

Since elastic potential energy U = $\frac{1}{2}Y{\in }^2, we can write U_A>U_B.$