Grams of bicarbonate vs realeased CO2?
Thank you for all the help, I've figured out the answer:

$$\ce{NaHCO3(aq) + HCl(aq) -&gt; NaCl(aq) + CO2(g) + H2O(l)}$$

Since we're given $\pu{1.60 g}$ of $\ce{CO2(g)}$ we can use the mass to calculate the number of moles with the equation: 

$$n = \frac{m}{M} = \frac{\pu{1.60 g}}{\pu{44.01 g mol-1}} = \pu{0.036355 mol}$$

Since it's a 1 : 1 ratio between $\ce{NaHCO3(aq)}$ and $\ce{CO2(g)}$, we can directly solve for the mass of $\ce{NaHCO3(aq)}$ using the same formula: 

$$n = \frac{m}{M} $$

$$m = n \times M = \pu{0.036355 mol} \times \pu{83.42 g mol-1} = \pu{3.0509116 g}$$ 

Round to two decimal places: $\pu{3.05 g}$