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).

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:

$$R = \boxed{\text{4.583 cm}}$$