The problem here is incomplete. Luckily, I found a similar problem from another website which is shown in the attached picture. So, the useful information here are the stoichiometric coefficients of the reaction and the initial amount of 7.49 g of S. The solution is as follows:
Molar mass of S = 32.06 g/mol
Molar mass of SO₃ = 80.06 g/mol
Theo yield of SO₃ = (7.49 g S)(1 mol S/32.06 g)(2 mol SO₃/2 mol S)(80.06 g SO₃/mol) = 18.7 g SO₃