The equivalent resistance between A and B of the network shown in the figure:

$$\frac{R_{AD}}{R_{DB}} = \frac{R}{3R} = \frac{1}{3}$$ ($$D$$ is the junction between $$9R$$ and $$3R$$)

$$\frac{R_{AC}}{R_{CB}} = \frac{2R}{6R} = \frac{1}{3}$$ ($$C$$ is the junction between $$9R$$ and $$2R$$)

Since $$\frac{R_{AD}}{R_{DB}} = \frac{R_{AC}}{R_{CB}} = \frac{1}{3}$$, the Wheatstone bridge is perfectly balanced.

$$R_{\text{top}} = R + 3R = 4R$$

$$R_{\text{bottom}} = 2R + 6R = 8R$$

$$R_{AB} = \frac{R_{\text{top}} \times R_{\text{bottom}}}{R_{\text{top}} + R_{\text{bottom}}}$$

$$R_{AB} = \frac{4R \times 8R}{4R + 8R} = \frac{32R^2}{12R} = \frac{8}{3}R$$