Let the cutting plane be passing through points B and C as shown

Height of cone = AD = H = 20 cm

Height of small cone which we get after cutting = AB = h_c

Let ‘r’ be the radius of small cone ∴ we have BC = r

‘R’ be radius of original cone which is to be cut ∴ we have DE = R

From figure consider ΔABC and ΔADE

∠ABC = ∠ADE = 90˚

∠BAC = ∠DAE …(common angle)

as two angles are equal by AA criteria we can say that

ΔABC∼ΔADE

Let V₁ be the volume of cone to be cut

Let V₂ be the volume of small cone which we get after cutting

Volume of cone = (1/3)π(radius)²(height) cm³

∴ V₁ = (1/3) × π × R² × h_c

∴ V₂ = (1/3) × π × r² × 20

Given is that the volume of small cone is (1/8) times the original cone

∴ V₂ = (1/8) V₁

∴ (1/3) × π × r² × 20 = (1/8) × (1/3) × π × R² × h_c

Using equation (i) we get

∴ h_c³ = 20³/8 cm

∴ h_c = 20/2 cm

∴ h_c = 10 cm

But we have to find the height from base i.e. we have to find BD from figure

∴ 20 = BD + h_c

∴ 20 = BD + 10

∴ BD = 10 cm

∴ 10 cm above base the section is made.