Taking it to the next level!

By symmetry, there are actually only two unique voltages in the circuit. Call these $V1$ and $V2$ . These are the voltages at the electrical nodes, which are the vertices of two regular pentagons. Write the nodal equations for a representative node from each pentagon, supposing that we apply a 1-volt DC source from B to A.

$3 V1 - 2 V2 = 1 \\ -2 V1 + 3 V2 = 0$

Solving gives $V1 = 0.6$ and $V2 = 0.4$ . The current flowing into B is:

$I = 5 (1 - V1) = 5 (1 - 0.6) = 2$

Since we applied 1 V and got 2 amps flowing out of the source, this corresponds to a total resistance of $0.5 \Omega$

Observe, that by symmetry the potentials at points $G,C,D,E,F$ are equal. So no current will flow from $GC,CD,DE,EF,FG$ .

So we can join the points $G,C,D,E,F$ .

Similarly, the potentials at $H,I,J,K,L$ are equal. So no current will flow from $HI,IJ,JK,KL,LH.$ So these points can also be joined.

So the circuit simplifies to:- Now from A to B, the resistances are in series, so adding them gives $R_{eq}=\boxed{\frac{1}{2} \Omega}$

its hard to write a solution at 3:00 A.M. but i will try my best.

Just like in case of a cube ,here also we use symmetry to eliminate the 10 resistors in the middle are they are connected across equipotential points.We are left with with a 5-10-5 parallel series chain which one easily fets R/2 as the equivalent resistsnce.