Answer
$E = 5.57\times 10^{-3}~N/C$
Work Step by Step
We can use an integral to find the charge within a sphere of radius $r = \frac{R}{2.00}$:
$q = \int_{0}^{R/2}~\rho~4\pi~r^2~dr$
$q = \int_{0}^{R/2}~\frac{(14.1~pC/m^3)~r}{R}~4\pi~r^2~dr$
$q = \int_{0}^{R/2}~\frac{(14.1~pC/m^3)~4\pi~r^3}{R}~dr$
$q = \frac{(14.1~pC/m^3)~4\pi~r^4}{4R} ~\Big \vert_{0}^{R/2}$
$q = \frac{(14.1~pC/m^3)~4\pi~(R/2.00)^4}{4R} -0$
$q = \frac{(14.1~pC/m^3)~4\pi~R^3}{64}$
$q = \frac{(14.1\times 10^{-12}~C/m^3)~(4\pi)~(0.0560~m)^3}{64}$
$q = 4.86\times 10^{-16}~C$
We can draw a Gaussian sphere with radius $r = \frac{R}{2.00}$ that is concentric with the sphere of charge.
We can find the magnitude of the electric field:
$\epsilon_0~\Phi = q$
$(\epsilon_0)(E)(4\pi~r^2) = q$
$E = \frac{q}{4\pi~\epsilon_0~r^2}$
$E = \frac{4.86\times 10^{-16}~C}{(4\pi)~(8.854\times 10^{-12}~F/m)~(0.0280~m)^2}$
$E = 5.57\times 10^{-3}~N/C$
