let x,y,z be numbers in order
y=2x,x=3z
(x+y+z)/3=20
(x+2x+(x/3))/9=20
solving
x=18
y=36
z=6


so largest is y=36

let the third no be x ,

then the first no is 3x ,
second no is 2*3x ie 6x

average is (3x+6x+x )/3 = 20

10x/3 = 20

x= 60/10 = 6

third no is 6 , second is 6x = 36, first no is 3x = 18
largest no is 36 i.e second no

third no be x
first no will be 3x
second no will be 6x
(x+3x+6x/3)=20
10x/3=20
10x=60
x=6

third no- x
first no- 3x
second no- 2*3x=6x

avg= (x+3x+6x)/3=20

thord nox=6
first no-18
second no-36
answer is 36

let 1st number be x 
thus 2nd number is 2x
let 3rd number be y
since 1st number is thrice the 3rd number, thus 
x= 3y

now : 1st no = 3y
      2nd no = 6y
      3rd no = y
so, (3y+6y+y)/3=20
              y = 6

since 2nd no is 6y, therefore 6y = 36
so the 2nd number is the greatest :D