Answer
$\displaystyle \frac{r_{b}}{r_{a}}=\frac{1}{3}$
Work Step by Step
Use the definition of resistivity: $ \qquad R=\displaystyle \rho(\frac{l}{A})$
Let
$l_{a}$=$l_{b} =l$ be the lengths of wires A and B, respectively,
$ A_{a}=r_{a}^{2}\pi,\quad A_{b}=r_{b}^{2}\pi$ be their cross-section areas.
Given $R_{b}=9R_{a}$, and $\rho$ being equal for both wires, we have:
$\displaystyle \rho(\frac{l}{A_{b}})=9\rho(\frac{l}{A_{a}})$
$\displaystyle \frac{1}{A_{b}}=\frac{9}{A_{a}}$
$\displaystyle \frac{A_{a}}{A_{b}}=9$
$\displaystyle \frac{A_{b}}{A_{a}}=\frac{1}{9}$
$\displaystyle \frac{r_{b}^{2}\pi}{r_{a}^{2}\pi}=\frac{1}{9}$
$\displaystyle \frac{r_{b}}{r_{a}}=\frac{1}{3}$
