Answer
$E = 5.99\times 10^3~N/C$
Work Step by Step
We can draw a Gaussian cylinder of radius $r = 2.00 R$ and length $L$ that has the same axis as the charged cylinder.
We can find the electric field at $r = 2.00 R$:
$\epsilon_0~\Phi = q_{enc}$
$\epsilon_0~E~2\pi~r~L = \lambda~L$
$E = \frac{\lambda}{2\pi~\epsilon_0~r}$
$E = \frac{\lambda}{4.00~\pi~\epsilon_0~R}$
$E = \frac{2.00\times 10^{-8}~C/m}{(4.00~\pi)~(8.854\times 10^{-12}~F/m)~(0.0300~m)}$
$E = 5.99\times 10^3~N/C$
