A circular disc of radius R carries surface charge density σ ( r )...

Consider a ring element of radius $r$, thickness in disc charge of element $dQ = \left(2\pi rdr\right)\sigma ⇒ dQ = 2\pir\sigma\left(dr\right)$

Total charge, $Q = ∫ 2\pir{\sigma}_{0}\left(1 - \dfrac{r}{R}\right)dr$

$Q = 2\pi{\sigma}_{0}\left(\dfrac{{r}^{2}}{2} - \dfrac{{r}^{3}}{3R}\right)$

${ϕ}_{0} = \dfrac{\text{charge upto}R}{{ϵ}_{0}} = 2\pi{\sigma}_{0}\left(\dfrac{{R}^{2}}{2} - \dfrac{{R}^{3}}{3R}\right)$

$ϕ = \dfrac{\text{charge upto}\dfrac{R}{4}}{{ϵ}_{0}} = 2\pi{\sigma}_{0}\left(\dfrac{{R}^{2}}{16 \times 2} - \dfrac{{R}^{3}}{64 \times 3R}\right)$

$\dfrac{{ϕ}_{0}}{ϕ} = \dfrac{\dfrac{1}{2} - \dfrac{1}{3}}{\dfrac{1}{32} - \dfrac{1}{192}} = \dfrac{1}{6} \times \dfrac{32 \times 6}{5} = 6 . 4$