First, we need to find the resistance of the copper wire as follows:
[tex]\begin{gathered} R=\frac{\rho L}{A} \\ where: \\ \rho=resistivity=1.7\times10^{-8}\Omega \\ L=length=8.433m \\ A=Area=\pi r^2=\pi(\frac{0.017}{2})^2=2.2698\times10^{-4}m^2 \end{gathered}[/tex]Therefore:
[tex]R\approx6.316\times10^{-4}\Omega[/tex]Now, using Ohm's law:
[tex]\begin{gathered} V=IR \\ so: \\ I=\frac{V}{R}=\frac{0.088}{6.316\times10^{-4}}\approx139.328A \end{gathered}[/tex]Answer:
139.328 A
