Answer
$E = 2.2\times 10^6~N/C$
Work Step by Step
The charge on the inner shell is $5.0\times 10^{-6}~C/m$
At a radius of $4.0~cm$, only the inner shell is enclosed within a cylindrical Gaussian surface.
We can use Equation (23-12) to find the electric field:
$E = \frac{\lambda}{2\pi~\epsilon_0~r}$
$E = \frac{5.0\times 10^{-6}~C/m}{(2\pi)~(8.854\times 10^{-12}~F/m)~(0.040~m)}$
$E = 2.2\times 10^6~N/C$
