Answer
$E = 13.5~N/C$
Work Step by Step
We can find the charge density from $r = 0$ to $r = 0.030~m$:
$\sigma = \int_{0}^{0.030~m}~Ar^2~dr$
$\sigma = \frac{Ar^3}{3}\Big\vert_{0}^{0.030~m}$
$\sigma = \frac{(2.5\times 10^{-6}~C/m^5)(0.030~m)^3}{3} - \frac{A(0)^3}{3}$
$\sigma = 2.25\times 10^{-11}~C/m^2$
We can find the linear charge density $\lambda$:
$\lambda = (2.25\times 10^{-11}~C/m^2)(1.0~m) = 2.25\times 10^{-11}~C/m$
We can use Equation (23-12) to find the electric field:
$E = \frac{\lambda}{2\pi~\epsilon_0~r}$
$E = \frac{2.25\times 10^{-11}~C/m}{(2\pi)~(8.854\times 10^{-12}~F/m)~(0.030~m)}$
$E = 13.5~N/C$
