A 9 Ω and an 18 Ω resistor are in parallel, and the pair is in series with a 4 Ω resistor, across…

Step 1. The pair. $\dfrac{9\ \Omega \times 18\ \Omega}{9\ \Omega + 18\ \Omega} = 6\ \Omega$.

Step 2. The total. $6\ \Omega + 4\ \Omega = 10\ \Omega$, so $I = \dfrac{25\ \text{V}}{10\ \Omega} = 2.5\ \text{A}$.

Step 3. The $9\ \Omega$ branch. The pair has $2.5\ \text{A} \times 6\ \Omega = 15\ \text{V}$, so $I_9 = \dfrac{15\ \text{V}}{9\ \Omega} = 1.67\ \text{A}$.