The excess pressure inside first soap bubble is three times that of a second soap bubble. The ratio of volumes of the first to second bubble

Given that the excess pressure $P_1$ in the first soap bubble is three times that in the second, $P_2$, we have $P_1 = 3P_2$.

Using the formula for excess pressure in a soap bubble, $P = \frac{4T}{R}$, we substitute to obtain $\frac{4T}{R_1} = 3\left(\frac{4T}{R_2}\right)$.

Cancelling the common factor $4T$ yields $\frac{1}{R_1} = \frac{3}{R_2}$, which simplifies to $R_2 = 3R_1$.

The volume of a sphere is $V = \frac{4}{3}\pi R^3$, so the volume ratio is $\frac{V_1}{V_2} = \frac{R_1^3}{R_2^3} = \left(\frac{R_1}{R_2}\right)^3$.

Substituting $\frac{R_1}{R_2} = \frac{1}{3}$ gives $\frac{V_1}{V_2} = \left(\frac{1}{3}\right)^3 = \frac{1}{27}$.

Final Answer: $\boxed{\text{C}}$