Answer
$\sigma = 3.8\times 10^{-8}~C/m^2$
Work Step by Step
Since the wire has a negative charge, the shell must have a positive charge which cancels out this negative charge.
We can consider the surface area of the shell when the length is 1.0 meter.
$\sigma~A = +3.6~nC$
$\sigma~(2\pi~r)~(1.0~m) = +3.6~nC$
$\sigma = \frac{+3.6~nC}{(2\pi)~(0.015~m)~(1.0~m)}$
$\sigma = 3.8\times 10^{-8}~C/m^2$
