A 200 W and a 50 W bulb, both for 220 V, are joined in series to a 220 V supply…

Step 1. Resistances. $\dfrac{(220\ \text{V})^2}{200\ \text{W}} = 242\ \Omega$ and $\dfrac{(220\ \text{V})^2}{50\ \text{W}} = 968\ \Omega$.

Step 2. Current. $I = \dfrac{220\ \text{V}}{1210\ \Omega} = 0.182\ \text{A}$.

Step 3. Powers. $(0.182\ \text{A})^2 \times 242\ \Omega = 8\ \text{W}$ and $(0.182\ \text{A})^2 \times 968\ \Omega = 32\ \text{W}$.

Why? In series the current is the same, so the larger resistance takes the larger power.