The circle of circumference $2\pi r$ has total resistance $2\pi r\lambda$. Between A and B (a quarter arc apart), the circle splits into two arcs in parallel:

$R_{\text{arc 1}}=\frac{\lambda \pi r}{2}$ (quarter arc), $R_{\text{arc 2}}=\frac{3\lambda \pi r}{2}$ (three-quarter arc).

The connecting wire AOB of length $2r$ has resistance $2r\lambda$, and it is in parallel with the two arcs.

Adding all three conductances in parallel:

$\frac{1}{R_{AB}}=\frac{2}{\lambda \pi r}+\frac{2}{3\lambda \pi r}+\frac{1}{2r\lambda }=\frac{1}{\lambda r}\left[\frac{2}{\pi }+\frac{2}{3\pi }+\frac{1}{2}\right]$

Common denominator $6\pi$:

$\frac{1}{R_{AB}}=\frac{1}{\lambda r}\cdot \frac{12+4+3\pi }{6\pi }=\frac{1}{\lambda r}\cdot \frac{16+3\pi }{6\pi }$

$\therefore R_{AB}=\frac{6\pi \lambda r}{16+3\pi }$