For this we need the atomic weights of Na, C, O and H:
Na=22.9893, C=12.0096, O=15.9994, H=1.008.
The molecular weight of Na₂CO₃ is 2×22.9893+12.0096+3×15.9994=105.9864.
In the reaction two molecules of NaOH are produced, and the atomic weight of the product is:
2(22.9893+15.9994+1.008)=79.9934.
The weight in grams of Na₂CO₃ is 1000. If the constant N relates the atomic weight to the weight in grams we have N=1000/105.9864=x/79.9934 where x is the weight of NaOH. So x=1000×79.9934/105.9864=754.75g approx.