Neutralisation stoichiometry requires matching the moles of acidic protons to the moles of hydroxide supplied. Oxalic acid dihydrate has the formula H2C2O4.2H2O with a molar mass of 126 g per mol, so the moles in 6.3 g equal 6.3/126 = 0.05 mol. Oxalic acid is diprotic, providing two ionisable protons per molecule, so the moles of acidic hydrogen are 2 × 0.05 = 0.10 mol, and complete neutralisation needs 0.10 mol of NaOH. The volume of 0.5 M NaOH that contains 0.10 mol is volume = moles/concentration = 0.10/0.5 = 0.20 L, which is 200 mL, the correct answer. The 100 mL option ignores the diprotic nature and treats the acid as monoprotic. The 50 mL option additionally mishandles the molar mass, and 400 mL doubles the requirement incorrectly. This calculation links NCERT carboxylic acid chemistry with volumetric analysis and is a typical JEE numerical. A common error is to forget the two waters of crystallisation when computing the molar mass, since omitting them gives 90 g per mol and a wrong mole count; the hydrated formula must always be used for a weighed sample of the dihydrate. Equivalently, the n-factor of oxalic acid here is two, so its equivalent mass is half its molar mass. As a sanity check, a diprotic acid needs twice the base of a comparable monoprotic acid, consistent with 200 mL.