Explanation

(a) $R = \rho\frac{l}{A}$ and mass $m = volume\ (V) \times density(d) = (A\ l)d$
Since wires have same material so $\rho$ and $d$ is same for both
Also they have same mass $\Rightarrow Al = constant \Rightarrow l \propto \frac{1}{A}$
$\Rightarrow \frac{R_{1}}{R_{2}} = \frac{l_{1}}{l_{2}} \times \frac{A_{2}}{A_{1}} = \left( \frac{A_{2}}{A_{1}} \right)^{2} = \left( \frac{r_{2}}{r_{1}} \right)^{4}$
$\Rightarrow \frac{34}{R_{2}} = \left( \frac{r}{2r} \right)^{4} \Rightarrow R_{2} = 544\ \Omega$