#### Answer

The required diameter of the aluminum wire is $0.64~mm$

#### Work Step by Step

We can write a general expression for the resistance:
$R = \frac{\rho~L}{A} = \frac{\rho~L}{\pi~r^2}$
We can write an expression for the resistance of the copper wire:
$R_c = \frac{\rho_c~L}{\pi~r_c^2}$
We can write an expression for the resistance of the aluminum wire:
$R_a = \frac{\rho_a~L}{\pi~r_a^2}$
Since the resistance must be equal, we can equate both expressions for the resistance:
$R_c = R_a$
$\frac{\rho_c~L}{\pi~r_c^2} = \frac{\rho_a~L}{\pi~r_a^2}$
$r_a^2 = \frac{\rho_a}{\rho_c}~r_c^2$
$r_a = \sqrt{\frac{\rho_a}{\rho_c}}~r_c$
$r_a = \sqrt{\frac{2.8\times 10^{-8}~\Omega~m}{1.7\times 10^{-8}~\Omega~m}}~(0.25~mm)$
$r_a = 0.32~mm$
Since the diameter is twice the radius, the required diameter of the aluminum wire is $0.64~mm$