Answer
$43.55\%$
Work Step by Step
Number of moles of $CO_2$:
$0.196\ g/44.001\ g/mol=0.00445\ mol$
From stoichiometry, the number of moles of sodium bicarbonate:
$0.00445\ mol\times 2_{Na_HCO_3}/1_{Na_3CO_3}=0.00891\ mol$
$0.00891\ mol\times 84.007\ g/mol=0.7484\ g$
Mass fraction:
$0.7484\ g/1.7184\ g=43.55\%$
