Given A1 = 12 cm² and A2 = 108 cm².

Let $A_1$ be the surface area of the smaller bottle and $A_2$ be the surface area of the larger bottle.
Given $A_1 = 12 \text{ cm}^2$ and $A_2 = 108 \text{ cm}^2$.
The ratio of the surface areas is:

$$\frac{A_1}{A_2} = \frac{12 \text{ cm}^2}{108 \text{ cm}^2} = \frac{1}{9}$$

For similar objects, the ratio of their surface areas is equal to the square of the ratio of their corresponding linear dimensions (scale factor, $k$).

$$\frac{A_1}{A_2} = k^2$$ $$\frac{1}{9} = k^2$$

To find $k$, take the square root of both sides:

$$k = \sqrt{\frac{1}{9}} = \frac{1}{3}$$

So, the ratio of the linear dimensions of the smaller bottle to the larger bottle is $\frac{1}{3}$.

For similar objects, the ratio of their volumes is equal to the cube of the ratio of their corresponding linear dimensions.
Let $V_1$ be the volume of the smaller bottle and $V_2$ be the volume of the larger bottle.
Given $V_2 = 810 \text{ cm}^3$.

$$\frac{V_1}{V_2} = k^3$$ $$\frac{V_1}{810 \text{ cm}^3} = \left(\frac{1}{3}\right)^3$$ $$\frac{V_1}{810 \text{ cm}^3} = \frac{1}{27}$$

Now, solve for $V_1$:

$$V_1 = \frac{1}{27} \times 810 \text{ cm}^3$$ $$V_1 = \frac{810}{27} \text{ cm}^3$$ $$V_1 = 30 \text{ cm}^3$$

The volume of the smaller bottle is $\text{30 cm}^3$.