ethylene glycol -> C2H6O2
20% of C2H6O2 by mass is present.
That means solution has 20 g of ethylene glycol and 80 g of water.
Now,
Molar mass of C2H6O2 = 12 x 2 + 1 x 6 + 2 x 16 = 62 g mol-1 .
Moles of C2H6O2 n1 = 20 / 62 = 0.322 moles
Moles of water n2 = 80 / 18 = 4.444 moles
Mole fraction of ethylene glycol =Moles of ethylene glycol / total no. of moles
= n1 / n1+n2
= 0.322 / (0.322 + 4.444) = 0.068