Answer
The net resistance of the circuit is $\frac{13R}{8}$
Work Step by Step
The two resistors in series between A and B have an equivalent resistance of 2R.
Let $R_1$ be the equivalent resistance of the three resistors between A and B.
$\frac{1}{R_1} = \frac{1}{2R}+\frac{1}{R}$
$\frac{1}{R_1} = \frac{1}{2R}+\frac{2}{2R}$
$\frac{1}{R_1} = \frac{3}{2R}$
Then $R_1 = \frac{2R}{3}$
$R_1$ and the resistor between B and C are in series with an equivalent resistance of $\frac{5R}{3}$
Let $R_2$ be the equivalent resistance of all the resistors between A and C.
$\frac{1}{R_2} = \frac{3}{5R}+\frac{1}{R}$
$\frac{1}{R_2} = \frac{3}{5R}+\frac{5}{5R}$
$\frac{1}{R_2} = \frac{8}{5R}$
Then $R_2 = \frac{5R}{8}$
Finally, $R_2$ and the last resistor are in series with an equivalent resistance of $\frac{5R}{8} + R$ which is $ \frac{13R}{8}$
The net resistance of the circuit is $\frac{13R}{8}$.
