Ratio of figure?

Here's a quick way to solve this without resorting to even knowing the formula for a cone.

Let's start by imagining we were comparing the small cone (height h) to the complete cone (height 3h).

1st dimension - length:
The ratio of the heights is h to 3h --> 1:3

2nd dimension - area:
If the problem asked about area, the ratio of areas (say the surface areas) would be the squares:
1² : 3² --> 1:9

3rd dimension - volume:
The ratio of their volumes will be in the ratio of their cubes.
1^3 : 3^3 --> 1:27

So the volume of the small cone can be thought of as 1 (volume) and the large cone is 27 (volumes).

But if you trim off the top (1 volume) you are left with 26 (volumes).

The ratio of the small cone to the truncated cone is:
1:26

Summary in all similar shapes:
Ratio of length --> a : b
Ratio of areas --> a² : b²
Ratio of volumes --> a^3 : b^3

Vh/V3h = (h/3h)³ = 1/27 = V / 27V

=> V2h = 27V - V = 26V

The searched ratio V/26V = 1/26
=========================

Vh/V3h = (h/3h)³
or, 1/27 = V / 27V
or, V2h = 27V - V = 26V

V/26V = 1/26

First find the volume of the large cone, then the volume of the small "upper" cone.
Subtract small from large. This gives volume of frustum.. Divide the small cone volume by the frustum volume. This is the ratio called for.

Right Cone:
V = (1/3)(pi)(r^2)(h)

Let the radius of the small cone be r, then the radius of the large cone is 2r, by similar triangles. The volume of the small cone is (1/3) pi r^2 h. The volume of the frustrum is (1/3) pi (3r)^2 * 3h - (1/3) pi r^2 h. Dividing each of these by (1/3) pi r^2 h (which doesn't change their ratio) gives their ratio as

1 : 26