Solution:

To find the resistance $R$ that must be placed in series with the bulb, let's analyze the problem step by step.

Given:

1. Power of the bulb ($P$) = 500 W,
2. Voltage rating of the bulb ($V_b$) = 100 V,
3. Supply voltage ($V_s$) = 200 V.

Step 1: Resistance of the bulb

The resistance of the bulb ($R_b$) can be calculated using the formula:

$$R_b = \frac{V_b^2}{P}.$$

Substitute the values:

$$R_b = \frac{100^2}{500} = \frac{10000}{500} = 20 \, \Omega.$$

Step 2: Total current through the circuit

The bulb is rated to draw 500 W at 100 V. Thus, the current through the bulb is:

$$I = \frac{P}{V_b}.$$

Substitute the values:

$$I = \frac{500}{100} = 5 \, \text{A}.$$

Step 3: Voltage drop across the series resistor

The total supply voltage is 200 V, and the bulb operates at 100 V. Therefore, the voltage drop across the series resistor $R$ is:

$$V_R = V_s - V_b.$$

Substitute the values:

$$V_R = 200 - 100 = 100 \, \text{V}.$$

Step 4: Resistance of the series resistor

Using Ohm's law, the resistance of the series resistor is:

$$R = \frac{V_R}{I}.$$

Substitute the values:

$$R = \frac{100}{5} = 20 \, \Omega.$$

Final Answer:

The resistance that must be placed in series with the bulb is:

$$\boxed{20 \, \Omega}.$$