Here's the solution to the problem: Let $R$ be the radius of the base of the whole cone (bottom radius of the frustum) and $r$ be the radius of the top of the frustum (radius of the small cone removed). Let $H$ be the height of the whole cone and $h$ be the height of the small cone removed. The height of the frustum is $h_f$. From the problem statement: The bottom diameter is thrice the top diameter. $2R = 3(2r) \implies R = 3r$. The height of the frustum is $h_f = 12$ m. We know that $h_f = H - h$. So, $H - h = 12$. Using similar triangles formed by the cross-section of the cone: $\frac{R}{H} = \frac{r}{h}$ Substitute $R = 3r$ into the ratio: $\frac{3r}{H} = \frac{r}{h}$ $3h = H$ Now we have a system of two equations: 1) $H - h = 12$ 2) $H = 3h$ Substitute (2) into (1): $3h - h = 12$ $2h = 12$ $h = 6$ m Now find $H$: $H = 3h = 3 \times 6 = 18$ m i) The height of the whole cone is: $$H = \boxed{\text{18 m}}$$ ii) Given that the volume of the whole cone is $V = 39,600 \text{ cm}^3$. We need to use the height of the whole cone in cm: $H = 18 \text{ m} = 18 \times 100 \text{ cm} = 1800 \text{ cm}$. The formula for the volume of a cone is $V = \frac{1}{3} \pi R^2 H$. We are given $\pi = \frac{22}{7}$. Substitute the known values into the volume formula: $$39600 = \frac{1}{3} \times \frac{22}{7} \times R^2 \times 1800$$ Simplify the equation: $$39600 = \frac{22 \times 1800}{3 \times 7} \times R^2$$ $$39600 = \frac{22 \times 600}{7} \times R^2$$ $$39600 = \frac{13200}{7} \times R^2$$ Now, solve for $R^2$: $$R^2 = \frac{39600 \times 7}{13200}$$ $$R^2 = \frac{396 \times 7}{132}$$ $$R^2 = 3 \times 7$$ $$R^2 = 21$$ Finally, find $R$ by taking the square root: $$R = \sqrt{21}$$ Calculate the value and round to four significant figures: $$R \approx 4.58257569...$$ $$R \approx 4.583 \text{ cm}$$ The radius of the whole cone, correct to four significant figures, is: $$R = \boxed{\text{4.583 cm}}$$