2 Na2S2O3 + AgBr = NaBr + Na3[Ag(S2O3)2]
The equation is balanced (I checked), so we can trust that we get one mole of NaBr for every 2 moles of Na2S2O3r. The molar ratio is 1/2, so if we start with 20.5 moles of NaBr, we'll get 10.25 moles of NaB. To find grams of that product, multiply 10.25 moles times the molar mass of NaBr.
NaBr has a molar mass of 102.9 grams/mole. To find grams:
(10.5 moles)*(401.1 grams/mole) = 1054.7 grams NaBr
Round to 1055 grams for 4 sig figs. I can only see 3 sig figs in the data (20.5 moles), but I put too much effort into it, so I'll leave it to you to truncate to 2 sig figs.