Since the conditions are at STP, 1 mole of gas (ideal) will occupy 22.4 liters. Find moles of Na present and then using the stoichiometry of the balanced e of oxygen and finally convert to liters.
moles Na present = 89 g x 1 mole/23 g = 3.9 moles Na
moles O2 = 3.9 mol Na x 1 mol O2/4 mol Na = 0.98 moles
volume = 0.98 mol x 22.4 L/mol = 22 L