Answer
The length should be $~~10.4~m$
Work Step by Step
We can find the required length:
$P = \frac{V^2}{R}$
$P = \frac{V^2}{\rho~L/A}$
$P = \frac{V^2~A}{\rho~L}$
$L = \frac{V^2~A}{\rho~P}$
$L = \frac{(100~V)^2~(2.60\times 10^{-6}~m^2)}{(5.00\times 10^{-7}~\Omega\cdot m)(5000~W)}$
$L = 10.4~m$
The length should be $~~10.4~m$
