The resistance of all the wires between any two adjacent dots is R. Then equivalent resistance between A and B as shown in figure is:

(1), (2) are in series and their effective is parallel to (3) $\therefore {\mathrm{R}}_{\mathrm{MN}}=\frac{\left(2\mathrm{R}\right)\mathrm{R}}{2\mathrm{R}+\mathrm{R}}=\frac{2\mathrm{R}}{3}$

Simillarly effective of (4), (5), (6) is ${\mathrm{R}}_{\mathrm{NB}}=\frac{2\mathrm{R}}{3}$

(7), (8), (9) is ${\mathrm{R}}_{\mathrm{BP}}=\frac{2\mathrm{R}}{3}$

(10), (11), (12) is ${\mathrm{R}}_{\mathrm{PO}}=\frac{2\mathrm{R}}{3}$

Now the remaining equivalent circuit is as shown in figure. The resistance joining M and O can be removed based on the Wheatstone bridge model :

In the left part :

${\mathrm{R}}^1=\frac{2\mathrm{R}}{3}+\frac{2\mathrm{R}}{3}+\mathrm{R}=\frac{7\mathrm{R}}{3}$

$\therefore {\mathrm{R}}_{\mathrm{AB}}=\frac{{\mathrm{R}}^1\times {\mathrm{R}}^1}{{\mathrm{R}}^1+{\mathrm{R}}^1}=\frac{{\mathrm{R}}^1}{2}=\frac{7\mathrm{R}}{6}$