Answer
$I=1.015\times 10^9A$
Work Step by Step
We know that:
$B=\frac{\mu_{\circ}I}{2r}$
This can be rearranged as
$I=\frac{2rB}{\mu_{\circ}}$
We plug in the known values to obtain:
$I=\frac{2\times 6.38\times 10^6\times 1\times 10^{-4}}{4\pi \times 10^{-7}}=1.015\times 10^9A$
