So whenever you are approaching stoichiometry questions you want to first figure out what's happening in the reaction. So in this case you know that they will be a displacement reaction that will form a barium salt:
Ba(NO3)2+ 2NaCl--> BaCl2 + 2NaNO3
So now that we have that formula and the molecular weight we can determine how much salt will be made. So here we convert the grams to moles
(42.3g Ba(NO3)2)*(1 mole/261.34g) = 0.16185 mol
In the molecular formula we know that 1 mole of Barium nitrate will create 1 mole of Barium chloride, so in this case (in a perfect world) you should get 0.16185 mole of barium chloride (208.23 g/mol) that we then have to convert to grams.
(0.16185 mol BaCl2) * ( 208.23 g/mol) = 33.7037 g of Barium Chloride (rounded to 3 significant digits = 33.7g)