Answer
The final resistance is $\frac{R}{3}$.
Work Step by Step
The old resistance is $R_{old}=\rho\frac{L}{A}$.
We are told that the length L triples.
We are told that the diameter triples, so the cross-sectional area A increases to 9A.
The new resistance is $R_{new}=\rho\frac{3L}{9A}=\frac{R_{old}}{3}$.
