A small soap bubble of radius 4 cm is trapped inside another bubble of radius 6 cm without any contact. Let P$$_2$$ be the pressure inside the inner bubble and P$$_0$$, the pressure outside the outer bubble. Radius of another bubble with pressure difference P$$_2$$ - P$$_0$$ between its inside and outside would be:

For a thin soap bubble the excess pressure inside over the outside is given by the Laplace formula:

$$\Delta P=\dfrac{4T}{r},$$

where $$T$$ is the surface tension and $$r$$ is the radius of the bubble. A soap bubble has two liquid surfaces, hence the factor 4 in the numerator.

Let us denote

$$P_0=\text{outside atmospheric pressure},$$

$$P_1$$ = pressure in the air layer between the two bubbles,

$$P_2=\text{pressure inside the smaller (inner) bubble}.$$

We have an outer bubble of radius $$6\ \text{cm}$$. Applying the formula to this outer bubble, the pressure difference between its inside and the atmosphere is

$$P_1-P_0=\dfrac{4T}{6\ \text{cm}}=\dfrac{2T}{3}.$$

Next, there is an inner bubble of radius $$4\ \text{cm}$$. The pressure difference between the inside of this inner bubble and the air immediately outside it (which is at pressure $$P_1$$) is

$$P_2-P_1=\dfrac{4T}{4\ \text{cm}}=T.$$

Now we add these two differences to obtain the total excess pressure inside the inner bubble relative to the atmosphere:

$$P_2-P_0=(P_2-P_1)+(P_1-P_0)=T+\dfrac{2T}{3}=\dfrac{5T}{3}.$$

Suppose we have a single soap bubble of unknown radius $$R$$ whose inside pressure exceeds the outside pressure by this same amount $$P_2-P_0$$. Using the Laplace formula once again, we write

$$P_2-P_0=\dfrac{4T}{R}.$$

Substituting the value already found for $$P_2-P_0$$, we get

$$\dfrac{4T}{R}=\dfrac{5T}{3}.$$

Dividing both sides by $$T$$ and cross-multiplying,

$$4\cdot 3 = 5R \;\;\Longrightarrow\;\; R=\dfrac{12}{5}\ \text{cm}=2.4\ \text{cm}.$$

Hence, the correct answer is Option A.