Pressure inside a soap bubble is greater than the pressure outside by an amount : (given : $$R$$ = Radius of bubble, $$S$$ = Surface tension of bubble)

We need to find the excess pressure inside a soap bubble over the outside pressure.

Key Concept: Excess pressure in a bubble with surface tension.

For a liquid drop (with one surface), the excess pressure inside due to surface tension is given by the Young-Laplace equation:

$$\Delta P_{drop} = \frac{2S}{R}$$

where $$S$$ is the surface tension and $$R$$ is the radius.

For a soap bubble: A soap bubble has two surfaces -- an inner surface and an outer surface. Both surfaces contribute to the excess pressure. Each surface contributes $$\frac{2S}{R}$$.

Therefore, the total excess pressure inside a soap bubble is:

$$\Delta P_{bubble} = \frac{2S}{R} + \frac{2S}{R} = \frac{4S}{R}$$

This is double the excess pressure of a single-surface liquid drop because the soap film has two air-liquid interfaces.

The correct answer is Option 4: $$\frac{4S}{R}$$.