Eleatric field due to complete disc $(R=2 a)$ at a distance $x$ and on its axis

$E_{1}=\frac{\sigma}{2 \varepsilon_{0}}\left[1-\frac{x}{\sqrt{R^{2}+x^{2}}}\right]$

$E_{1}=\frac{\sigma}{2 \varepsilon_{0}}\left[1-\frac{h}{\sqrt{4 a^{2}+h^{2}}}\right]$

$ = \frac{\sigma }{{2{\varepsilon _0}}}\left[ {1 – \frac{h}{{2a}}} \right]$

$\left[ \begin{gathered}
  {\text{ here }}x = h{\text{ }} \hfill \\
  {\text{and }},R = 2a \hfill \\ 
\end{gathered}  \right]$

Similarly, electric field due to disc $(R=a)$

$E_{2}=\frac{\sigma}{2 \varepsilon_{0}}\left(1-\frac{h}{a}\right)$

Electric field due to given disc $E=E_{1}-E_{2}$

$\frac{\sigma}{2 \varepsilon_{0}}\left[1-\frac{h}{2 a}\right]-\frac{\sigma}{2 \varepsilon_{0}}\left[1-\frac{h}{a}\right]-\frac{\sigma h}{4 \varepsilon_{0} a}$

Hence, $c=\frac{\sigma}{4 a \varepsilon_{0}}$