The answer is 0.667.
The combustion of acetylene can be written as
2C_2H_2 + 5O_2 -> 4CO_2 + 2H_2O
The balanced chemical reaction shows a 2:5 mole ratio for C_2H_2 and O_2, a 1:2 ratio for C_2H_2 and CO_2, and a 1:1 ratio for C_2H_2 and H_2O.
Now, for simplicity, let us assume that we have 1 mol of C_2H_2 to start with ( the equivalence of 26 grams ); given the fact that the oxygen is in excess, acetylene will always act as a limiting reagent.
So, one would get 2 moles of CO_2 and 1 mole of H_2O in the products, for a total of n_T = 2 + 1 = 3 moles of product.
Therefore, the mole fraction of CO_2 would be
n_(CO_2) = (n_(CO_2)/n_T) = 2/3 = 0.667
In this case, 20% excess O_2 would mean that you started with 3 moles of O_2 instead of 2.5, but I don't see how this would be relevant other than determining that O_2 is not a limiting reagent.
