Interpret the network

The given resistor network is the standard infinite ladder in which each repeating section contains:

• a resistor $R$ in series along the main branch, and
• after that node, a resistor $2R$ shunted to the lower rail, and then the same structure repeats indefinitely.

Let the equivalent resistance between $A$ and $B$ be $X$.

Use self-similarity of the infinite ladder

After the first section, the remaining part of the network is again identical to the whole network. Hence the resistance of the remaining part is also $X$.

So the first section looks like:

• one series resistor $R$,
• then a parallel combination of $2R$ and $X$.

Therefore, $X = R + \left(2R \parallel X\right).$

Write the parallel combination

$2R \parallel X = \frac{2R\,X}{2R+X}.$

Hence, $X = R + \frac{2RX}{2R+X}.$

Solve the equation

Multiply both sides by $(2R+X)$: $X(2R+X) = R(2R+X) + 2RX.$

Expand both sides: $2RX + X^2 = 2R^2 + RX + 2RX.$

$2RX + X^2 = 2R^2 + 3RX.$

Rearranging, $X^2 - RX - 2R^2 = 0.$

Solve the quadratic

$X = \frac{R \pm \sqrt{R^2 + 8R^2}}{2} = \frac{R \pm 3R}{2}.$

So, $X = 2R \quad \text{or} \quad X = -R.$

Since resistance cannot be negative, $X = 2R.$

Match with the options

The equivalent resistance should be $2R$.

But this is not present in the listed options. The stored answer is option $B = \frac{8}{3}R$, which does not match the self-similar infinite-ladder result.

Conclusion

Based on the standard interpretation of the infinite repeating network, the equivalent resistance is $\boxed{2R}.$