To solve this problem, we need to use the concept of normality and the stoichiometry of the reaction between oxalic acid and sodium hydroxide. Oxalic acid dihydrate (C2H2O4.2H2O) reacts with sodium hydroxide (NaOH) in a 1:2 molar ratio because oxalic acid is a diprotic acid, meaning each molecule of oxalic acid can donate two protons.

First, we need to calculate the number of moles of oxalic acid in the solution:

The molar mass of oxalic acid dihydrate (C2H2O4.2H2O) is calculated as follows:

C: 12.01 g/mol * 2 = 24.02 g/mol

H: 1.01 g/mol * 4 = 4.04 g/mol

O: 16.00 g/mol * 6 = 96.00 g/mol

Total: 126.06 g/mol

Given mass of oxalic acid dihydrate = 6.3 g

Moles of oxalic acid dihydrate = $$\frac{\text{mass}}{\text{molar mass}} = \frac{6.3 g}{126.06 g/mol} = 0.05 \text{ mol}$$

Since the solution is made up to 250 mL, the molarity (M) of this solution is calculated as follows:

$$M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} = \frac{0.05 \text{ mol}}{0.25 \text{ L}} = 0.2 \text{ M}$$

When 10 mL of this solution reacts with NaOH, the volume of NaOH required to completely neutralize the solution can be found using the stoichiometry of the reaction:

The reaction between oxalic acid (H2C2O4) and NaOH is represented by the balanced chemical equation:

$$H_2C_2O_4 + 2NaOH \rightarrow Na_2C_2O_4 + 2H_2O$$

This indicates that 1 mole of oxalic acid reacts with 2 moles of NaOH. Therefore, the moles of oxalic acid in 10 mL of the solution can be found as:

$$\text{Moles of oxalic acid} = 0.2 M * 0.01 L = 0.002 \text{ mol}$$

Since each mole of oxalic acid requires 2 moles of NaOH for complete neutralization:

$$\text{Moles of NaOH required} = 0.002 \text{ mol of oxalic acid} * 2 = 0.004 \text{ mol of NaOH}$$

The normality (N) of NaOH solution is given as 0.1N. Because NaOH is monovalent, its normality is equal to its molarity in this context. Hence, the volume of NaOH solution required can be calculated using the formula:

$$V = \frac{\text{moles of NaOH}}{\text{Normality of NaOH solution}} = \frac{0.004 \text{ mol}}{0.1 N} = 0.04 L = 40 mL$$

Therefore, the correct answer is Option A: 40mL.