Assuming the equivalent resistance is between opposite corners. There are 3 resistors between attached to each node. Inject 3A into the first node so 1A goes to each leg. Use KCL to further divide or add the currents as it moves to successive nodes. You’ll find the voltage drop (sum of currents in path between nodes times 1 ohm) is 2.5V or 5/2V. The current out of terminal node has to be 3A since that is what went in. Add up the currents in the terminal node and it had better be 3A or you did something wrong. So equivalent resistance is 2.5V/3A or (5/2)/(6/2 )=5/6