Answer
$E = 3.1\times 10^6~N/C$
This maximum value occurs at a radial distance of $~~R = 5.0~cm~~$ from the center of the pipe.
Work Step by Step
We can draw a Gaussian cylinder of length $L$ with the axis along the same axis as the cylindrical pipe.
We can find the magnitude of the electric field at a distance $r$ from the pipe center where $r \lt R$:
$\epsilon_0~\Phi = q_{enc}$
$(\epsilon_0)~(E)~(2\pi~r~L) = \rho~\pi~r^2~L$
$(\epsilon_0)~(E)~(2) = \rho~r$
$E = \frac{\rho~r}{2~\epsilon_0}$
We can see that $E$ increases with increasing $r$
The maximum value of $E$ occurs when $r = R$
We can find the maximum value of $E$:
$E = \frac{\rho~r}{2~\epsilon_0}$
$E = \frac{(1.1\times 10^{-3}~C/m^3)(0.050~m)}{(2)(8.854\times 10^{-12}~F/m)}$
$E = 3.1\times 10^6~N/C$
This maximum value occurs at a radial distance of $~~R = 5.0~cm~~$ from the center of the pipe.
