Answer
$90.\ m$
Work Step by Step
Mass of the ingot in g: $57\ kg \times 10^3\ g/kg = 57\times10^3\ g$
Given the density of $8.96\ g/cm^3$, the volume is: $V=m/\rho=6.4\times10^3\ cm^3$
With the volume of the wire given by $V=\pi r^2L$, and the radius (half of the diameter) being $4.75\ mm = 0.475\ cm$, the length is:
$L=9.0\times10^3\ cm=90.\ m$
