Step 1: Formula for excess pressure in liquid drop.
\[
\Delta P = \frac{2T}{r}
\]
Step 2: Relation between excess pressures.
Given:
\[
\Delta P_1 = 3\Delta P_2
\]
\[
\frac{2T}{r_1} = 3\cdot\frac{2T}{r_2}
\]
Step 3: Finding ratio of radii.
\[
r_2 = 3r_1
\]
Step 4: Ratio of surface areas.
\[
\frac{A_1}{A_2} = \frac{4\pi r_1^2}{4\pi r_2^2} = \frac{r_1^2}{(3r_1)^2} = \frac{1}{9}
\]
Step 5: Conclusion.
The ratio of their surface areas is $1:9$.