Let's find the focal length of the lens in air and in the liquid. We will use the lens maker's formula:

$$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

where $$f$$ is the focal length, $$\mu$$ is the refractive index of the lens material, and $$R_1$$ and $$R_2$$ are the radii of curvature of the lens surfaces.

Since the lens is convex, both surfaces have the same radius of curvature (positive), so $$R_1 = R_2 = 20 \mathrm{~cm}$$.

First, let's find the focal length of the lens in air:

$$\frac{1}{f_\text{air}} = (1.8 - 1)\left(\frac{1}{20} - \frac{1}{20}\right) = 0.8\left(\frac{1}{20}\right)$$

$$f_\text{air} = \frac{1}{0.8\left(\frac{1}{20}\right)} = 25 \mathrm{~cm}$$

Now, let's find the focal length of the lens in the liquid. The relative refractive index of the lens with respect to the liquid is:

$$\mu_\text{rel} = \frac{1.8}{1.5} = 1.2$$

$$\frac{1}{f_\text{liquid}} = (1.2 - 1)\left(\frac{1}{20}\right)$$

$$f_\text{liquid} = \frac{1}{0.2\left(\frac{1}{20}\right)} = 100 \mathrm{~cm}$$

The power of a lens is given by:

$$P = \frac{1}{f}$$

Now, we can find the ratio of the power of the lens in air to its power in the liquid:

$$\frac{P\text{air}}{P\text{liquid}} = \frac{f\text{liquid}}{f\text{air}} = \frac{100}{25} = 4$$

So, the ratio of the power of the lens in air to its power in the liquid is $$x:1$$, where $$x = 4$$.