A cell has an emf of 1.58 V and an internal resistance of 0.45 Ω…

Step 1. Current. The total resistance is $3.6\ \Omega + 0.45\ \Omega = 4.05\ \Omega$. So $I = \dfrac{1.58\ \text{V}}{4.05\ \Omega} = 0.390\ \text{A}$.

Step 2. Terminal voltage. The voltage lost inside the cell is $Ir = 0.390\ \text{A} \times 0.45\ \Omega = 0.176\ \text{V}$. So $V = 1.58\ \text{V} - 0.176\ \text{V} = 1.404\ \text{V}$.

Step 3. Power in the bulb. $P = VI = 1.404\ \text{V} \times 0.390\ \text{A} = 0.548\ \text{W}$.