Answer
$d = 4.6~mm$
Work Step by Step
We can find the resistance of the aluminum rod:
$R = \frac{\rho~L}{A}$
$R = \frac{(2.75\times 10^{-8}~\Omega\cdot m)(1.3~m)}{(5.2\times 10^{-3}~m)^2}$
$R = 1.3\times 10^{-3}~\Omega$
Let $d$ be the diameter of the copper rod.
We can find the required length of $d$:
$R = \frac{\rho~L}{A}$
$R = \frac{\rho~L}{\pi~(d/2)^2}$
$R = \frac{4~\rho~L}{\pi~d^2}$
$d^2 = \frac{4~\rho~L}{\pi~R}$
$d = \sqrt{\frac{4~\rho~L}{\pi~R}}$
$d = \sqrt{\frac{(4)~(1.69\times 10^{-8}~\Omega\cdot m)~(1.3~m)}{(\pi)~(1.3\times 10^{-3}~\Omega)}}$
$d = 4.6\times 10^{-3}~m$
$d = 4.6~mm$
