(B) The graph shows the variation of current $(I)$ on the $y$-axis with respect to voltage $(V)$ on the $x$-axis.
According to Ohm's law, $V = IR$, which implies $I = (1/R)V$.
The slope of the $I-V$ graph is given by $\text{slope} = I/V = 1/R$.
From the graph, the slope of line $X$ is greater than the slope of line $Y$ $(\text{slope}_X > \text{slope}_Y)$.
Since $\text{slope} = 1/R$, this means $1/R_X > 1/R_Y$, which implies $R_X < R_Y$.
Resistance $(R)$ is related to resistivity $(\rho)$ by the formula $R = \rho(L/A)$.
Given that both wires have the same length $(L)$ and the same diameter (and thus the same cross-sectional area $A$), the resistance is directly proportional to the resistivity $(R \propto \rho)$.
Since $R_Y > R_X$, it follows that the resistivity of wire $Y$ is greater than the resistivity of wire $X$ $(\rho_Y > \rho_X)$.