1. Condition for maximum average power in a series LCR circuit

For a series LCR circuit, the average power is $P = VI\cos\phi = I^2R$ and it becomes maximum when the impedance is minimum.

In a series LCR circuit, $Z = \sqrt{R^2 + (X_L - X_C)^2}$ This is minimum when $X_L = X_C$ So, for maximum average power, the circuit must be in resonance.

2. Use the resonance condition

Given: $X_L = \frac{70}{11}\,\Omega$ At resonance, $X_C = X_L = \frac{70}{11}\,\Omega$

Also, $X_C = \frac{1}{\omega C}$ where $\omega = 2\pi f$

Given: $f = 50\,\text{Hz}, \quad \pi = \frac{22}{7}$ So, $\omega = 2\times \frac{22}{7} \times 50 = \frac{2200}{7}\,\text{rad/s}$

3. Calculate capacitance

Using $X_C = \frac{1}{\omega C}$ we get $C = \frac{1}{\omega X_C}$

Substitute values: $C = \frac{1}{\left(\frac{2200}{7}\right)\left(\frac{70}{11}\right)}$

Now simplify: $\left(\frac{2200}{7}\right)\left(\frac{70}{11}\right)= \frac{2200\times 70}{7\times 11}$

$= \frac{2200\times 10}{11} = 2000$

Thus, $C = \frac{1}{2000}\,\text{F}$

Convert into microfarads: $\frac{1}{2000}\,\text{F} = 5\times 10^{-4}\,\text{F}$

Since $1\,\mu\text{F} = 10^{-6}\,\text{F}$

$C = 5\times 10^{-4} \times 10^6 = 500\,\mu\text{F}$

4. Final answer

$\boxed{500}$

The derived answer matches the stored correct answer.