In the given network of resistances, the equivalent resistance across terminals A and B is estimated to be

The middle node joining the two $$3\ \Omega$$ resistors is also joined to the point between the two $$6\ \Omega$$ resistors. So between A and this node there is $$3 \parallel 6 = 2\ \Omega$$, and between the node and B there is another $$3 \parallel 6 = 2\ \Omega$$, giving $$4\ \Omega$$ in series. This $$4\ \Omega$$ is in parallel with the $$12\ \Omega$$ resistor directly across A and B, so $$R_{AB} = \frac{4 \times 12}{16} = 3\ \Omega$$.