Answer
$E = 2.23\times 10^{-2}~N/C$
Work Step by Step
In part (a), we found that the total charge is $~~q = 7.78\times 10^{-15}~C$
We can draw a Gaussian sphere with radius $r = R$ that is concentric with the sphere of charge.
We can find the magnitude of the electric field:
$\epsilon_0~\Phi = q$
$(\epsilon_0)~(E)(4\pi~r^2) = q$
$E = \frac{q}{4\pi~\epsilon_0~r^2}$
$E = \frac{7.78\times 10^{-15}~C}{(4\pi)~(8.854\times 10^{-12}~F/m)~(0.0560~m)^2}$
$E = 2.23\times 10^{-2}~N/C$
