Here we are assuming that " $\ell$ " is very large just for the sake of symmetry. Outside cylinder will have zero electric field inside, so the flux generated on the plate will be due to inner cylinder only in sections AB and CD , as section be will be at that place where electric field is zero.

Flux through element will be =

$$ \begin{aligned} \mathrm{d} \phi & =\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dS}} \\ \mathrm{~d} \phi & =\frac{2 \mathrm{k} \lambda}{\mathrm{r}} \mathrm{dy} \cdot \ell \cdot \cos \theta .........(1) \end{aligned} $$

from figure we can say that

$$ \begin{aligned} & \cos \theta=\frac{R}{r} \Rightarrow r=R \sec \theta \\ & \tan \theta=\frac{y}{R} \Rightarrow y=R \tan \theta \\ & \Rightarrow d y=R \sec ^2 \theta d \theta \\ & d \phi=\frac{2 k \lambda}{R \sec \theta} R \cdot \sec ^2 \theta \cdot \ell \cdot \cos \theta d \theta \\ & d \phi=2 k \lambda \ell d q \end{aligned} $$

$\begin{aligned} & \int_0^{\phi_{\mathrm{AB}}} \mathrm{d} \phi=2 \mathrm{k} \lambda \ell \int_{\pi / 4}^{\pi / 3} \mathrm{~d} \theta \\ & \phi_{\mathrm{AB}}=2 \mathrm{k} \lambda \ell\left[\frac{\pi}{3}-\frac{\pi}{4}\right] \\ & \phi_{\mathrm{AB}}=2 \mathrm{kQ}\left[\frac{\pi}{12}\right] \\ & \phi_{\mathrm{AB}}=2 \times \frac{1}{4 \pi \epsilon_0 \epsilon_{\mathrm{r}}} \mathrm{Q}\left[\frac{\pi}{12}\right] \\ & \phi_{\mathrm{AB}}=\frac{\mathrm{Q}}{120 \epsilon_0} \\ & \phi_{\text {plate }}=\phi_{\mathrm{AB}}+\phi_{\mathrm{BC}}+\phi_{\mathrm{CD}} \quad\left(\phi_{\mathrm{CD}}=\phi_{\mathrm{AB}}\right) \\ & \phi_{\text {plate }}=\frac{\mathrm{Q}}{60 \epsilon_0}\end{aligned}$