Calculate the volume of a regular pyramid with a base side length of 9 cm and a slant height of 18 cm.

The base is a square with side length $s = 9$ cm.

$$\text{Base Area} (A_b) = s^2$$ $$A_b = 9^2$$ $$A_b = 81 \text{ cm}^2$$

For a square base, the distance from the center of the base to the midpoint of a side (apothem, $a$) is half the side length.

$$a = \frac{s}{2}$$ $$a = \frac{9}{2} = 4.5 \text{ cm}$$

The slant height ($l$), the perpendicular height ($h$), and the apothem ($a$) form a right-angled triangle. We can use the Pythagorean theorem:

$$l^2 = h^2 + a^2$$ $$18^2 = h^2 + (4.5)^2$$ $$324 = h^2 + 20.25$$ $$h^2 = 324 - 20.25$$ $$h^2 = 303.75$$ $$h = \sqrt{303.75}$$

To simplify $\sqrt{303.75}$:

$$303.75 = \frac{30375}{100} = \frac{1215}{4}$$ $$h = \sqrt{\frac{1215}{4}} = \frac{\sqrt{1215}}{\sqrt{4}} = \frac{\sqrt{81 \times 15}}{2} = \frac{9\sqrt{15}}{2} \text{ cm}$$

$$V = \frac{1}{3} A_b h$$ $$V = \frac{1}{3} (81) \left(\frac{9\sqrt{15}}{2}\right)$$ $$V = 27 \left(\frac{9\sqrt{15}}{2}\right)$$ $$V = \frac{243\sqrt{15}}{2}$$

To get a numerical value:

$$V \approx \frac{243 \times 3.873}{2}$$ $$V \approx \frac{941.079}{2}$$ $$V \approx 470.54 \text{ cm}^3$$

The volume of the regular pyramid is $\boxed{\frac{243\sqrt{15}}{2} \text{ cm}^3 \approx 470.54 \text{ cm}^3}$.