Answer
$i=12.7A$
Work Step by Step
The magnetic dipole moment is given as:
$\mu=NiA$
or $\mu=Ni \pi r^2$
This can be rearranged as:
$i=\frac{\mu}{N\pi r^2}$
We plug in the known values to obtain:
$i=\frac{2.30}{160\times 3.1416\times (0.0190)^2}$
$i=12.7A$
