Note that nodes A,B,C are at the same potential. So are D,E and F,G,H,I. Thus we have 3 sets of equipotential nodes: i){A,B,C} ii){D,E} iii){F,G,H,I}.
Next, the resistances across these equipotential nodes are in parallel. Thus, the resistance R between A,D and the resistance R between B,E are in parallel with their equivalent resistance as 0.5R. Similarly, the resistors R between D,I E,H and E,F respectively are in parallel with their equivalent resistance R3R3.
On merging the equipotential nodes and writing these quivalent resistances, the circuit look as below:
This circuit is solvable, yielding equivalent resistance as equal to
Req=5R11
