First we need to convert the amounts of NaHCO3 and acetic acid into moles and then compare the two.
5 g NaHCO3 * (mole/84 g) = 0.06 moles NaHCO3
90 mL AcOH * (1.05 g/mL) * (mole/60.05 g) = 1.57 moles AcOH
From the balanced chemical equation you gave in your question, we know that both species react together in a 1 to 1 molar fashion. As a result, we have many more moles of acetic acid present than NaHCO3, so sodium bicarbonate will be the limiting reagent. Thus, 0.06 moles of NaHCO3 will produce 0.06 moles of CH3COONa (sodium acetate). Let's convert this to grams.
0.06 moles AcONa * (82.03 g/mole) = 4.92 g AcONa
So 5 g of sodium bicarbonate will theoretically produce 4.92 g of sodium acetate. So the theoretical yield is 4.92 g.