In the frustrum of a cone, the bottom diameter is thrice the top diameter. (a) If the height of the frustrum is 12 cm, calculate the height of the whole cone. (b) Given that the volume of the whole cone is 39,600 cm³, find, correct to four significant figures, the radius. [Take π = 22/7 ]

Let $R$ be the radius of the base of the whole cone and $r$ be the radius of the top of the frustum (which is the base of the small cone removed).
Given that the bottom diameter is thrice the top diameter, we have $2R = 3(2r)$, which simplifies to $R = 3r$.

Let $H$ be the height of the whole cone and $h_s$ be the height of the small cone that was cut off. The height of the frustum is $h_f = H - h_s = 12$ cm.
Due to similar triangles formed by the whole cone and the small cone, the ratio of their radii is equal to the ratio of their heights:

Substitute $R = 3r$:

Substitute $h_s = \frac{H}{3}$ into the frustum height equation $H - h_s = 12$:

The height of the whole cone is $\boxed{\text{18 cm}}$.

The volume of a cone is given by $V = \frac{1}{3}\pi R^2 H$.

Given $V = 39,600 \text{ cm}^3$, $H = 18 \text{ cm}$ (from part a), and $\pi = \frac{22}{7}$.

Rounding to four significant figures:

The radius of the whole cone is $\boxed{\text{45.83 cm}}$.