Answer
$\phi^{\prime}_E=-1.1\times 10^{-4}\frac{N.m^2}{C}$
Work Step by Step
The flux through the rim of radius $a=0.11m$ can be calculated as:
$\phi_E=EA$
$\implies \phi_E=E\pi a^2$
We plug in the known values to obtain:
$\phi_E=(3.0\times 10^{-3})(3.1416)(0.11)^2=1.1\times 10^{-4}\frac{N.m^2}{C}$
The flux through the net $(\phi^{\prime}_E)$ is opposite to that of rim:
$\implies \phi^{\prime}_E=-\phi_E=-1.1\times 10^{-4}\frac{N.m^2}{C}$
