1. Find the lamp resistance

The lamp consumes power $P=500\,\text{W}$ and the voltage across it is $V_R=100\,\text{V}$.

Since the lamp is treated as a resistive element, $P=V_R I$ So, $I=\frac{P}{V_R}=\frac{500}{100}=5\,\text{A}$

Hence its resistance is $R=\frac{V_R}{I}=\frac{100}{5}=20\,\Omega$

2. Use series RC voltage relation

The supply voltage is $V=200\,\text{V}$.

In a series $RC$ circuit, the resistor voltage and capacitor voltage are perpendicular in phase, so $V^2=V_R^2+V_C^2$ Thus, $200^2=100^2+V_C^2$ $V_C^2=40000-10000=30000$ $V_C=100\sqrt{3}\,\text{V}$

3. Find capacitive reactance

Since $V_C=IX_C$ we get $X_C=\frac{V_C}{I}=\frac{100\sqrt{3}}{5}=20\sqrt{3}\,\Omega$

4. Use capacitive reactance formula

$X_C=\frac{1}{\omega C}$ with $\omega=2\pi f=2\pi(50)=100\pi$ So, $20\sqrt{3}=\frac{1}{100\pi C}$ $C=\frac{1}{100\pi\cdot 20\sqrt{3}}=\frac{1}{2000\pi\sqrt{3}}\,\text{F}$

5. Convert to microfarads

$C=\frac{10^6}{2000\pi\sqrt{3}}\,\mu\text{F}$ $C=\frac{500}{\pi\sqrt{3}}\,\mu\text{F}$

Given $\pi\sqrt{3}\approx 5$, $C\approx \frac{500}{5}=100\,\mu\text{F}$

6. Phase angle

For completeness, $\cos\varphi=\frac{V_R}{V}=\frac{100}{200}=\frac12$ so $\varphi=60^\circ$ This is consistent with the above result.

Therefore, the required capacitance is $\boxed{100}$