#### Answer

The diameter must be increased by a factor of 1.26

#### Work Step by Step

We can write an expression for the resistance in a wire.
$R = \frac{\rho~L}{A} = \frac{\rho~L}{\pi~r^2}$
We can write an expression for the resistance in the copper wire.
$R_c = \frac{\rho_c~L}{\pi~r_c^2}$
We can write an expression for the resistance in the aluminum wire.
$R_a = \frac{\rho_a~L}{\pi~r_a^2}$
Since the resistance is the same in both wires, we can equate the two expressions to find the factor that the radius must be increased.
$R_c = R_a$
$\frac{\rho_c~L}{\pi~r_c^2} = \frac{\rho_a~L}{\pi~r_a^2}$
$r_a^2 = \frac{\rho_a~r_c^2}{\rho_c}$
$r_a = \sqrt{\frac{\rho_a}{\rho_c}}~r_c$
$r_a = \sqrt{\frac{2.73\times 10^{-8}~\Omega~m}{1.72\times 10^{-8}~\Omega~m}}~r_c$
$r_a = 1.26~r_c$
Since the radius must be increased by a factor of 1.26, the diameter must be increased by a factor of 1.26