To determine the volume of 0.1N NaOH required to completely neutralize 10 ml of the given solution of oxalic acid dehydrate, we need to follow a series of steps.

1. Calculate the molarity of the oxalic acid solution:
- Given mass of oxalic acid dehydrate = 6.3g
- Molar mass of oxalic acid dehydrate (C2H2O4·2H2O) = 126.07 g/mol
- Moles of oxalic acid dehydrate = mass/molar mass = 6.3g/126.07 g/mol = 0.05 mol
- Volume of the solution = 250 ml = 0.25 L
- Molarity (M) of the solution = moles/volume = 0.05 mol/0.25 L = 0.2 M

2. Write the balanced chemical equation for the reaction between oxalic acid (H2C2O4) and sodium hydroxide (NaOH):
H2C2O4 + 2NaOH -> Na2C2O4 + 2H2O

3. Determine the stoichiometry of the reaction:
From the balanced equation, we can see that 1 mole of oxalic acid reacts with 2 moles of sodium hydroxide.

4. Calculate the moles of oxalic acid reacting in 10 ml of the solution:
Moles of oxalic acid = molarity × volume = 0.2 mol/L × 0.010 L = 0.002 mol

5. Use the stoichiometry to determine the moles of sodium hydroxide required:
Since the stoichiometric ratio is 1:2, the moles of NaOH required will be double the moles of oxalic acid reacting.
Moles of NaOH required = 2 × 0.002 mol = 0.004 mol

6. Calculate the volume of 0.1N NaOH needed to provide 0.004 moles of NaOH:
Molarity (M) = moles/volume
0.1 N = 0.1 mol/L
Volume of 0.1N NaOH = moles/molarity = 0.004 mol/0.1 mol/L = 0.04 L = 40 ml

Therefore, the volume of 0.1N NaOH required to completely neutralize 10 ml of the given solution of oxalic acid dehydrate is 40 ml.