Resistors of 12 Ω, 24 Ω and 24 Ω are in parallel across a 6 V battery…

Step 1. The combination. $\dfrac{1}{R_p} = \dfrac{1}{12\ \Omega} + \dfrac{1}{24\ \Omega} + \dfrac{1}{24\ \Omega} = \dfrac{4}{24\ \Omega}$, so $R_p = 6\ \Omega$.

Step 2. The current. $I = \dfrac{6\ \text{V}}{6\ \Omega} = 1\ \text{A}$.