Answer
1.64
Work Step by Step
$R_{1}=R_{2}$
$\implies \rho_{Al}\times\frac{l_{1}}{A_{1}}=\rho_{Cu}\times\frac{l_{2}}{A_{2}}$
Given that $l_{1}=l_{2}=l$
Therefore,
$\rho_{Al}\times\frac{l}{A_{1}}=\rho_{Cu}\times\frac{l}{A_{2}}$
$\implies \frac{A_{1}}{A_{2}}=\frac{\rho_{Al}}{\rho_{Cu}}=\frac{2.82\times10^{-8}\,\Omega\cdot m}{1.72\times10^{-8}\,\Omega\cdot m}=1.64$
