1. Pressure inside a soap bubble

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

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

where $T$ is the surface tension.

So, for a single soap bubble of radius $r$, the internal pressure is

$P_{\text{single}} = P_0 + \frac{4T}{r}$

where $P_0$ is atmospheric pressure.

2. Pressure inside the outer bubble

The larger soap bubble has radius $6\,\text{cm}$. Its internal pressure is

$P_1 = P_0 + \frac{4T}{6} = P_0 + \frac{2T}{3}$

3. Pressure inside the smaller bubble formed inside it

The smaller bubble of radius $3\,\text{cm}$ is formed inside the larger bubble, so the pressure outside the small bubble is not atmospheric pressure; it is $P_1$.

Hence pressure inside the smaller bubble is

$P_2 = P_1 + \frac{4T}{3}$

Substitute $P_1$:

$P_2 = \left(P_0 + \frac{2T}{3}\right) + \frac{4T}{3} = P_0 + 2T$

So, the internal pressure of the smaller bubble is

$P_2 = P_0 + 2T$

4. Equate with pressure inside a single soap bubble of radius $r$

Given that this equals the internal pressure of another single soap bubble of radius $r$:

$P_0 + \frac{4T}{r} = P_0 + 2T$

Cancelling $P_0$:

$\frac{4T}{r} = 2T$

$\frac{4}{r} = 2$

$r = 2\,\text{cm}$

5. Final answer

$\boxed{2}$

The derived answer matches the stored correct answer.