Explanation

(a) According to Gauss’s law total flux coming out of a closed surface enclosing charge $q$ is given by
$\phi = \oint_{}^{}{\overrightarrow{E}.d\overrightarrow{S}} = \frac{q}{\varepsilon_{0}}$
From this expression, it is clear that total flux linked with a closed surface only depends on the enclosed charge and independent of the shape and size of the surface.
$\phi = \oint_{}^{}{\overrightarrow{E}.d\overrightarrow{S}} = \frac{q}{\varepsilon_{0}} = 20\ Vm$ given
This $\frac{q}{\varepsilon_{0}}$ is constant as long as the enclosed charge is constant
$\Rightarrow$ The flux over a concentric sphere of radius $20\ cm = 20Vm$