1. Pressure inside a soap bubble

For a soap bubble of radius $R$, the excess pressure inside it is

$\Delta P = \frac{4T}{R}$

where $T$ is the surface tension.

So for the two bubbles:

• For radius $R_1 = 2\text{ cm}$, $P_1 - P_0 = \frac{4T}{2} = 2T$

• For radius $R_2 = 4\text{ cm}$, $P_2 - P_0 = \frac{4T}{4} = T$

Here $P_0$ is atmospheric pressure.

Thus,

$P_1 = P_0 + 2T, \qquad P_2 = P_0 + T$

2. Pressure difference across the common surface

The pressure difference between the two bubbles is

$P_1 - P_2 = (P_0 + 2T) - (P_0 + T) = T$

3. Common surface is a soap film

The common surface between the two bubbles is again a soap film, so the pressure difference across it is

$P_1 - P_2 = \frac{4T}{r}$

where $r$ is the radius of curvature of the common surface.

Therefore,

$\frac{4T}{r} = T$

$r = 4\text{ cm}$

4. Final answer

The radius of curvature of the common surface is

$\boxed{4}$