Area of the base = 11 × 4 = 44 m^{2}
And volume of the cone = 11 × 20 = 220 m^{3}
1/3 × πR^{2}h = 220 m^{3 }….(i)
Area of the base = πR^{2}
πR^{2} = 44
R^{2} = 44/22 × 7;
R^{2} = 14
R = √14 ……(ii)
By equation (i) and (ii), we get
1/3 × 22/7 × √14 × √14 × h = 220
h = 220 × 3/22 × 2
h = 30/2 = 15 cm