The molar mass of the citric acid is 192 g/mol.
The balanced chemical equation for the neutralization reaction that takes place between sodium hydroxide, NaOH, and citric acid, C6H8O7, looks like this
Notice the 1:3 mole ratio that exists between citric acid and sodium hydroxide; what this tells you is that, for every mole of citric acid, you need 3 times more moles of sodium hydroxide for the reaction to take place.
Since you know the molarity and the volume of the NaOH you've used, you can calculate how many moles of NaOH reacted
nNaOH=0.223 M⋅36.10⋅10−3L=0.00805 moles
Now use the aforementioned mole ratio to see how many moles of citric acid were present in 0.515 g
0.00805moles NaOH⋅1 mole citric acid3moles NaOH=0.00268 moles citric acid
Now simply divide the mass of citric acid given by the number of moles it contained to get the compound's molar mass
MM=mn=0.515 g0.00268 moles=192.16 g/mol
Rounded to three sig figs, the number of sig figs given for 0.515 g, the answer will be