Total Current supplied to the circuit by the battery is

4A

In the circuit resistors of 2$\Omega$ and $6\Omega$ are in parallel its equivalent is $\frac{2 \times 6}{2+6}\Omega = 1.5\Omega$

$1.5 \Omega$ is in series with resistor of $1.5 \Omega$, Its equivalent is $3 \Omega$.

Now both resistances of 3 $\Omega$ are in parallel so equivalent resistance is $R_{eq} = 1.5 \Omega$

Current $I = \frac{6 V}{1.5 \Omega} = 4A$