Since the potential difference between B and D is zero, no current flows through the diagonal wire.
This implies the circuit is a balanced Wheatstone bridge.
The two resistors in each arm are in series: 2Ω + 2Ω and 3Ω + 3Ω.
Net resistance in each arm = 4Ω and 6Ω respectively.
Now, total resistance between A and C is: $R_{total} = \dfrac{4 \cdot 6}{4 + 6} = \dfrac{24}{10} = 2.4\ \Omega$
Given voltage = 12V, using Ohm’s law: $I = \dfrac{V}{R} = \dfrac{12}{6} = 2$ A (as per configuration shown)