Answer
The net charge on the shell is $~~3.8\times 10^{-8}~C$
Work Step by Step
We can use an integral to sum each layer of surface area between $r = 4.0~cm$ and $r = 6.0~cm$, multiplied by the charge density at each layer.
$Q = \int_{0.040}^{0.060}~(\frac{b}{r})(4\pi~r^2~dr)$
$Q = \int_{0.040}^{0.060}~4\pi~b~r~dr$
$Q = 2\pi~b~r^2 \Big \vert_{0.040}^{0.060}$
$Q = (2\pi)~(3.0~\mu C/m^2)~[(0.060~m)^2-(0.040~m)^2]$
$Q = 3.8\times 10^{-8}~C$
The net charge on the shell is $~~3.8\times 10^{-8}~C$
