To solve this problem, we need to understand how the resistance of a wire changes when its dimensions are altered.$\\$The resistance $R$ of a wire is given by the formula:$\\ R = \frac{\rho L}{A} \\$where:$\\$• $\rho$ is the resistivity of the material,$\\$• $L$ is the length of the wire, and$\\$• $A$ is the cross-sectional area of the wire.$\\$Initially, the resistance of the wire is $R. \\$When the wire is melted and stretched to $n$ times its original length, the new length $L'$ is:$\\ L' = nL \\$Since the volume of the wire remains constant, we have:$\\ A \cdot L = A' \cdot L' \\$where $A'$ is the new cross-sectional area.$\\$Substitute $L' = nL$ into the volume equation:$\\ A \cdot L = A' \cdot nL \\$Solve for $A': \\ A' = \frac{A}{n} \\$Now, substitute $L'$ and $A'$ into the resistance formula to find the new resistance $R': \\ R' = \frac{\rho L'}{A'} \\ R' = \frac{\rho nL}{\frac{A}{n}} \\ R' = \frac{\rho n^2 L}{A} \\ R' = n^2 \cdot \frac{\rho L}{A} \\ R' = n^2 R \\$Therefore, the new resistance of the wire is $n^2 R. \\$This corresponds to Option 2.