Top radius of frustum (R) = 20 cm

The volume of a frustum ($V_f$) is given by:

$$V_f = \frac{1}{3}\pi h_f (R^2 + Rr + r^2)$$

$$V_f = \frac{1}{3}\pi (24) (20^2 + (20)(15) + 15^2)$$ $$V_f = 8\pi (400 + 300 + 225)$$ $$V_f = 8\pi (925)$$ $$V_f = 7400\pi$$

The volume of the bucket is $\boxed{\text{7400}\pi \text{ cm}^3}$.

Let $H$ be the height of the original cone and $h_s$ be the height of the small cone removed from the top.
Using similar triangles, the ratio of radii is equal to the ratio of heights:

$$\frac{R}{H} = \frac{r}{h_s}$$

We know $H = h_f + h_s$. Substitute this into the ratio:

$$\frac{R}{h_f + h_s} = \frac{r}{h_s}$$ $$R h_s = r (h_f + h_s)$$ $$R h_s = r h_f + r h_s$$ $$R h_s - r h_s = r h_f$$ $$h_s (R - r) = r h_f$$ $$h_s = \frac{r h_f}{R - r}$$

Substitute the values: $R=20$ cm, $r=15$ cm, $h_f=24$ cm.

$$h_s = \frac{15 \times 24}{20 - 15}$$ $$h_s = \frac{360}{5}$$ $$h_s = 72 \text{ cm}$$

$$H = h_f + h_s$$ $$H = 24 + 72$$ $$H = 96 \text{ cm}$$

The volume of a cone ($V_c$) is given by:

$$V_c = \frac{1}{3}\pi R^2 H$$

Substitute the values: $R=20$ cm and $H=96$ cm.

$$V_c = \frac{1}{3}\pi (20)^2 (96)$$ $$V_c = \frac{1}{3}\pi (400) (96)$$ $$V_c = \pi (400) (32)$$ $$V_c = 12800\pi$$

The volume of the original cone is $\boxed{\text{12800}\pi \text{ cm}^3}$.