Answer
The wire is made of nichrome.
Work Step by Step
We can find the resistivity of the wire:
$\rho = \frac{E}{J}$
$\rho = \frac{E}{I/A}$
$\rho = \frac{E~A}{I}$
$\rho = \frac{E~\pi~r^2}{I}$
$\rho = \frac{(0.0075~V/m)~(\pi)~(0.50\times 10^{-3}~m)^2}{3.9\times 10^{-3}~A}$
$\rho = 1.5\times 10^{-6}\Omega~m$
This value for the resistivity is closest to the resistivity of nichrome. We can conclude that the wire is made of nichrome.
