A point charge + Q is placed just outside an imaginary...

Every point on circumference of flat surface is at equal distance from point charge 

Hence circumference is equipotential.

Flux passing through curved surface = - flux passing through flat surface.

${\left(dϕ\right)}_{throughthering} = Ecos \theta . dA = \dfrac{1}{4\pi{\in}_{0}}\dfrac{Q}{\left(\sqrt{{r}^{2} + {R}^{2}}\right)}\dfrac{R}{\sqrt{{R}^{2} + {r}^{2}}} . 2\pi rdr$

$∴ ∫ dϕ = \dfrac{QR}{4\pi{\in}_{0}}2\pi\mathop{∫}_{0}^{R}\dfrac{rdr}{{\left({R}^{2} + {r}^{2}\right)}^{\dfrac{3}{2}}} = \dfrac{q}{2{\in}_{0}}\left(1 - \dfrac{1}{\sqrt{2}}\right)$

$∴$ Flux through curved surface $= - \dfrac{q}{2{\in}_{0}}\left(1 - \dfrac{1}{\sqrt{2}}\right)$

Note: Flux through surface can be calculated using concept of solid angle.

$Ω = 2\pi\left(1 - cos \theta\right) = 2\pi\left(1 - \dfrac{1}{\sqrt{2}}\right)$

$∴$ Solid angle subtended$= 2\pi\left(1 - \dfrac{1}{\sqrt{2}}\right)$

$ϕ$ for  $4\pi$ solid angle $= \dfrac{q}{{\in}_{0}}$

$∴ ϕ$ for  $2\pi\left(1 - \dfrac{1}{\sqrt{2}}\right)$ solid angle $= \dfrac{q}{4\pi{\in}_{0}} . 2\pi\left(1 - \dfrac{1}{\sqrt{2}}\right)$

$= \dfrac{q}{2{\in}_{0}}\left(1 - \dfrac{1}{\sqrt{2}}\right)$