An electricity and magnetism problem by Soumava Pal

U missed to mention that as the distribution of current is symmetric you can plug up the node and then proceed

Since the circuit is symmetrical, we can just do it by setting up a relation with the resistance between the second inner triangle and the outermost triangle. So we replace the second inner triangle that is the medial triangle of the medial triangle of the outermost triangle by a resistance $R_{AB}/2$ . Now let $R_{AB}$ =r, then if the resistance of AB is R, we have,

$r=R(R+\frac{Rr/2}{R+{\frac{r}{2}}})(R+R+\frac{Rr/2}{R+{\frac{r}{2}}})^{-1}$

Solving this equation yields,

$r=R{\frac{\sqrt{7}-1}{3}}$

Here R=1. So answer is simply, $\frac{\sqrt{7}-1}{3}$