b
The equivalent structure is shown in figure. Here the capacitance between inner and middle cylinder is equal to equivalent capacitance when $C_{1}$ and $C_{2}$ are in parallel.
$C_{1}=\frac{2 \pi \in_{0}}{\ln \left(\frac{2 R}{R}\right)}=\frac{2 \pi \in_{0}}{\ln 2}$
$C_{2}=\frac{2 \pi \in_{0}}{\ln \left(\frac{2 \sqrt{2 R}}{2 R}\right)}=\frac{4 \pi \in_{0}}{\ln 2}$
$C_{e q}=C_{1}+C_{2}=\frac{2 \pi \in_{0}}{l n 2}+\frac{4 \pi \in_{0}}{l n 2}=\frac{6 \pi \in_{0}}{l n 2}$
