Introduction:
To determine the required inductance to save the electric bulb, we need to analyze the given information about the electric bulb and the AC supply.

Given Information:
- Power of the electric bulb: 100 W
- Voltage of the electric bulb: 300 V
- AC supply voltage: 500 V
- AC supply frequency: 150/Pi Hz

Calculating the Current:
Using the power formula, we can calculate the current flowing through the electric bulb.

P = VI

Where,
P = Power (in watts)
V = Voltage (in volts)
I = Current (in amperes)

Substituting the given values:

100 W = 300 V × I

I = 100 W / 300 V

I = 0.33 A

Calculating the Impedance:
The impedance of the circuit can be calculated using the formula:

Z = V / I

Where,
Z = Impedance (in ohms)
V = Voltage (in volts)
I = Current (in amperes)

Substituting the given values:

Z = 500 V / 0.33 A

Z = 1515.15 Ω

Calculating the Inductive Reactance:
The inductive reactance (XL) can be calculated using the formula:

XL = 2πfL

Where,
XL = Inductive Reactance (in ohms)
f = Frequency (in hertz)
L = Inductance (in henries)

Substituting the given values:

XL = 2π × (150/Pi) Hz × L

Calculating the Required Inductance:
Now, we can equate the impedance (Z) to the inductive reactance (XL) to find the required inductance.

Z = XL

1515.15 Ω = 2π × (150/Pi) Hz × L

Simplifying the equation:

L = (1515.15 Ω) / (2π × (150/Pi) Hz)

L = (1515.15 Ω × Pi × (150/Pi) Hz) / (2π)

L = (1515.15 × 150) / (2 × Pi)

L ≈ 14334.18 H

Conclusion:
The required inductance to save the electric bulb is approximately 14334.18 H (henries). This calculation ensures that the impedance of the circuit matches the inductive reactance, allowing the electric bulb to operate properly and be protected from damage.