Answer
$\phi=7.1\frac{N.m^2}{C}$
Work Step by Step
We know that:
$\phi=\frac{q}{\epsilon_{\circ}}$
$\implies \phi=\frac{\sigma \pi r^2}{\epsilon_{\circ}}$
$\sigma$ and $\pi r^2$ represent surface charge density and area respectively.
We plug in the known values to obtain:
$\phi=\frac{8.0\times 10^{-9}\times 3.1416\times (0.050)^2}{8.85\times 10^{-12}}=7.1\frac{N.m^2}{C}$
