A soap bubble of the radius 3 cm is formed inside another soap bubble of radius 6 cm. The radius of an equivalent soap bubble that has the same excess pressure as inside the smaller bubble with respect to the atmospheric pressure is _________ cm.

The excess (gauge) pressure inside a soap bubble over the pressure of the medium that surrounds the bubble is given by the formula $$\Delta P = \dfrac{4T}{r}$$, where $$T$$ is the surface tension of the soap film and $$r$$ is the radius of the bubble. This factor 4 appears because a soap bubble has two free surfaces, an inner and an outer one.

We have two concentric bubbles. The larger bubble, whose radius is $$r_2 = 6 \, \text{cm}$$, is in direct contact with the atmosphere. The smaller bubble, of radius $$r_1 = 3 \, \text{cm}$$, is lodged inside the larger one and is therefore surrounded not by the atmosphere but by the air that already exists inside the larger bubble.

First, we calculate the excess pressure inside the larger bubble over atmospheric pressure. Using the stated formula,

$$\Delta P_{\text{large}} = \dfrac{4T}{r_2} = \dfrac{4T}{6} = \dfrac{2T}{3}.$$ Thus the pressure just inside the wall of the larger bubble is higher than atmospheric pressure by $$\dfrac{2T}{3}$$.

Next, we determine the excess pressure inside the smaller bubble relative to the air that surrounds it (which is the air inside the larger bubble). Again applying the formula,

$$\Delta P_{\text{small w.r.t. large}} = \dfrac{4T}{r_1} = \dfrac{4T}{3}.$$

Therefore, the total excess pressure of the air at the centre of the smaller bubble over the atmospheric pressure is obtained by simple addition of these two pressure rises, because pressures add algebraically along the same line of action:

$$\Delta P_{\text{small w.r.t. atmosphere}} = \Delta P_{\text{large}} + \Delta P_{\text{small w.r.t. large}} = \dfrac{2T}{3} + \dfrac{4T}{3} = \dfrac{6T}{3} = 2T.$$

We are asked for the radius $$r_{\text{eq}}$$ of a single isolated soap bubble that would have exactly this same excess pressure $$2T$$ over the atmosphere. Setting the general expression equal to this value, we write

$$\dfrac{4T}{r_{\text{eq}}} = 2T.$$

Dividing both sides by $$T$$ and then by 2, we carry out the straightforward algebraic steps:

$$\dfrac{4}{r_{\text{eq}}} = 2 \;\;\Longrightarrow\;\; r_{\text{eq}} = \dfrac{4}{2} = 2 \, \text{cm}.$$

So, the answer is $$2 \text{ cm}$$.