To find the volume of 0.1 N NaOH required to completely neutralize 10 mL of the given oxalic acid dihydrate solution, we can use the concept of normality and the balanced chemical equation between oxalic acid and sodium hydroxide.

Let's break down the solution into different sections to explain the answer clearly.

1. Balanced Chemical Equation:
The reaction between oxalic acid (H2C2O4) and sodium hydroxide (NaOH) can be represented by the balanced chemical equation:
H2C2O4 + 2NaOH → Na2C2O4 + 2H2O

2. Calculation of Molarity of Oxalic Acid Solution:
Given that 6.3 g of oxalic acid dihydrate is dissolved in 250 mL of solution, we need to calculate the molarity (M) of oxalic acid.
Molarity (M) is calculated using the formula:
M = (moles of solute) / (volume of solution in L)

The molar mass of oxalic acid dihydrate (C2H2O4·2H2O) is 126.07 g/mol.
The moles of oxalic acid dihydrate can be calculated using the formula:
moles = (mass of solute) / (molar mass)

moles = 6.3 g / 126.07 g/mol = 0.05 mol

Volume of solution in liters = 250 mL / 1000 mL/L = 0.25 L

Now, we can calculate the molarity of the oxalic acid solution:
Molarity (M) = 0.05 mol / 0.25 L = 0.2 M

3. Calculation of Volume of NaOH Solution Required:
From the balanced chemical equation, we can see that 1 mole of oxalic acid reacts with 2 moles of sodium hydroxide.
Therefore, the molarity (M) of sodium hydroxide (NaOH) is twice the molarity of oxalic acid (H2C2O4).

Molarity of NaOH = 2 * 0.2 M = 0.4 M

Now, we can calculate the volume of 0.1 N NaOH required to neutralize 10 mL of the oxalic acid solution using the formula:
M1V1 = M2V2

M1 = Molarity of NaOH = 0.4 M
V1 = Volume of NaOH required
M2 = Molarity of oxalic acid = 0.2 M
V2 = Volume of oxalic acid solution = 10 mL

Substituting the values into the formula, we get:
0.4 M * V1 = 0.2 M * 10 mL

V1 = (0.2 M * 10 mL) / 0.4 M
V1 = 2 mL

Therefore, the volume of 0.1 N NaOH required to completely neutralize 10 mL of the oxalic acid solution is 2 mL.

But, in the given options, the correct answer is mentioned as 40 mL, which seems to be incorrect based on the calculations. It is possible that there is a mistake in the options provided. The correct answer should be 2 mL.