An 8 Ω and a 24 Ω resistor are in parallel, and this pair is in series with a 4 Ω resistor…

Step 1. The pair. $R_p = \dfrac{8\ \Omega \times 24\ \Omega}{8\ \Omega + 24\ \Omega} = 6\ \Omega$.

Step 2. Total. $6\ \Omega + 4\ \Omega = 10\ \Omega$, so $I = \dfrac{18\ \text{V}}{10\ \Omega} = 1.8\ \text{A}$.

Step 3. Across the pair. $1.8\ \text{A} \times 6\ \Omega = 10.8\ \text{V}$.

Step 4. The branches. $\dfrac{10.8\ \text{V}}{8\ \Omega} = 1.35\ \text{A}$ and $\dfrac{10.8\ \text{V}}{24\ \Omega} = 0.45\ \text{A}$.