The excess pressure inside a soap bubble is determined by the formula:

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

where $ \Delta \mathrm{P} $ is the excess pressure, $ \mathrm{T} $ is the surface tension of the soap film, and $ \mathrm{R} $ is the radius of the bubble.

Given that the excess pressure inside bubble A is half that inside bubble B, we have:

$ \Delta \mathrm{P}_{\mathrm{A}} = \frac{1}{2} \Delta \mathrm{P}_{\mathrm{B}} $

This implies:

$ \frac{R_{\mathrm{A}}}{R_{\mathrm{B}}} = \frac{\Delta \mathrm{P}_{\mathrm{B}}}{\Delta \mathrm{P}_{\mathrm{A}}} = 2 $

Now, considering the volumes of the bubbles, the relationship between volume and radius for a sphere is given by:

$ V = \frac{4}{3} \pi R^3 $

Thus, the ratio of the volumes of bubbles A and B is:

$ \frac{V_{\mathrm{A}}}{V_{\mathrm{B}}} = \left( \frac{R_{\mathrm{A}}}{R_{\mathrm{B}}} \right)^3 = 2^3 = 8 $

Therefore, the volume of bubble A is $ n = 8 $ times the volume of bubble B.