Answer
$I=16A$
Work Step by Step
For the magnetic fields to cancel out, the magnitude of the magnetic field from the wire must equal the magnetic field from the Earth. Use equation 29-4 from page 837 to find the magnetic field. $$B=\frac{\mu_o I}{2\pi R}$$ Solving for current $I$ yields $$I=\frac{2\pi RB}{\mu_o}$$ Substitute known values of $B=39 \times 10^{-6}T$, $R=8.0cm=0.080m$, and $\mu_o=4\pi \times 10^{-7} H/m$ yields a current of $$I=\frac{2\pi (0.080m)(39\times 10^{-6}T)}{4\pi \times 10^{-7}H/m}=16A$$
