The current in the circuit is 0.5 A..
Given data:
The voltage of the battery is V=12 V.
The resistances are,

The voltage of the battery is V=12 V.
The equivalent resistance of the given series connection of resistances can be calculated as,

Applying the Ohm's law to calculate the current in the circuit as,

Thus, the current in the circuit is 0.5 A.