The number of air molecules per cm$$^3$$ is increased from $$3 \times 10^{19}$$ to $$12 \times 10^{19}$$. The ratio of collision frequency of air molecules before and after the increase in number respectively is:

The collision frequency of a gas molecule (number of collisions per unit time for a single molecule) is given by

$$f = \sqrt{2}\;\pi d^2 n \bar{v}$$

where $$d$$ is the molecular diameter, $$n$$ is the number density, and $$\bar{v}$$ is the mean speed.

Since the temperature remains constant, both $$\bar{v}$$ and $$d$$ remain the same (because mean speed depends only on temperature and molecular mass). So $$f \propto n$$.

The number density increases from $$n_1 = 3 \times 10^{19}\;\text{cm}^{-3}$$ to $$n_2 = 12 \times 10^{19}\;\text{cm}^{-3}$$. The ratio of collision frequency before and after is then

$$\frac{f_1}{f_2} = \frac{n_1}{n_2} = \frac{3 \times 10^{19}}{12 \times 10^{19}} = \frac{1}{4} = 0.25$$

Hence, the correct answer is 0.25.