To find the volume of 0.1N NaOH solution required to neutralize completely 10 mL of a 6.3 g oxalic acid solution, we need to first determine the number of moles of oxalic acid present in the solution.

1. Calculate the molar mass of oxalic acid:
The molar mass of oxalic acid (H2C2O4) is:
(2 × 1.01 g/mol) + (2 × 12.01 g/mol) + (4 × 16.00 g/mol) = 90.03 g/mol

2. Calculate the number of moles of oxalic acid present in the solution:
Given mass of oxalic acid = 6.3 g
Number of moles = mass / molar mass = 6.3 g / 90.03 g/mol = 0.07 mol

3. Calculate the volume of 0.1N NaOH solution required to neutralize 1 mole of oxalic acid:
The balanced chemical equation for the reaction between oxalic acid and sodium hydroxide (NaOH) is:
H2C2O4 + 2NaOH → Na2C2O4 + 2H2O

From the equation, we can see that 1 mole of oxalic acid reacts with 2 moles of NaOH. Therefore, the volume of 0.1N NaOH solution required to neutralize 1 mole of oxalic acid is 2 L.

4. Calculate the volume of 0.1N NaOH solution required to neutralize 0.07 moles of oxalic acid:
Volume of 0.1N NaOH solution = (0.1 mol/L) × (0.07 mol) = 0.007 L = 7 mL

5. Determine the volume of 0.1N NaOH solution required to neutralize completely 10 mL of the oxalic acid solution:
The volume of 0.1N NaOH solution required to neutralize completely 10 mL of the oxalic acid solution is proportional to the number of moles of oxalic acid in the solution. Therefore, the volume of 0.1N NaOH solution required is:
(7 mL / 1 mole) × (0.07 moles) = 0.49 mL

However, it is important to note that the options provided in the question are in mL, so we need to convert the volume from L to mL:
0.49 mL × 1000 = 490 mL

Therefore, the correct answer is option C: 40 mL.