First you have to calculate how many mols of NaOH are in the 25.0 cm3. As one dm3 is equal to 10cm*10cm*10cm=1000 cm3, there are 0.0016 mol/cm3 and therefore 25cm3*0.0016 mol/cm3=0.04 mol NaOH.
According to the stochiometrics of the reaction a maximum yield of 0.5*0.04=0.02 mol Na2SO4 can be formed. Multiplying by the molar mass gives the mass: 0.02mol*142.04g/mol=2.84 g
