(A) We know that $V = IR$,which implies $R = V/I$. The slope of the $V-I$ graph represents the resistance $R$ of the wire.
From the given graph,the slope of line $A$ is greater than the slope of line $B$,therefore,the resistance of wire $A$ $(R_A)$ is greater than the resistance of wire $B$ $(R_B)$.
We know that the resistance of a wire is given by the formula $R = \rho \frac{l}{A}$,where $\rho$ is the resistivity,$l$ is the length,and $A$ is the cross-sectional area.
Since both wires are made of the same material,their resistivity $\rho$ is the same. Since they are of equal thickness,their cross-sectional area $A$ is also the same.
Thus,$R \propto l$. Since $R_A > R_B$,it follows that the length of wire $A$ is greater than the length of wire $B$.