$$
R=\frac{\rho l}{A}, R_{A}=\frac{\rho l_{A}}{\pi r_{A}^{2}}, R_{B}=\frac{\rho l_{B}}{\pi r_{B}^{2}}
$$
Since the radius of \(A\) is three times that of \(B\) and the length of \(A\) is half of \(B, r_{A}=3 r_{B}\) and \(l_{A}=1 / 2 l_{B}\)
The ratio of resistance of \(A\left(R_{A}\right)\) to
that of \(B\left(R_{B}\right): \frac{R A}{R B}\)
The ratio of resistance of \(A\left(R_{A}\right)\) to
that of \(B\left(R_{B}\right): \frac{R_{A}}{R_{B}}\)
$$
\begin{aligned}
\frac{\rho \times \frac{1}{2} l_{B}}{\pi\left(3 r_{B}\right)^{2}} \div \frac{\rho \times l_{B}}{\pi r_{B^{2}}} &=\frac{\rho \times \frac{1}{2} V_{B}}{\pi \times 9 r / B^{2}} \times \frac{\pi r / B^{2}}{\rho \times 1 / B} \\
\frac{1}{2 / 9} &=\frac{9}{2} .
\end{aligned}
$$
\(\rho\) remains constant because the material of \(A\) is the same as that of \(B\)