An aqueous solution of oxalic acid dihydrate contains its $$6.3$$ g in $$250$$ ml. The volume of $$0.1$$ N NaOH required to completely neutralize $$10$$ ml of this solution

Molar mass of anhydrous oxalic acid, $$H_2C_2O_4$$, is $$24 + 2 + 64 = 90\text{ g mol}^{-1}$$.
Oxalic acid dihydrate contains two extra water molecules, so

$$M\,(\text{H}_2\text{C}_2\text{O}_4\cdot 2\text{H}_2\text{O}) = 90 + 2\times18 = 126\text{ g mol}^{-1}$$

Oxalic acid is dibasic (it can furnish two $$H^+$$ ions).
Hence, equivalent weight $$E$$ is

$$E = \frac{M}{\text{basicity}} = \frac{126}{2} = 63\text{ g eq}^{-1}$$

Mass of the acid taken = $$6.3\text{ g}$$.
Number of equivalents present:

$$\text{equivalents} = \frac{\text{mass}}{E} = \frac{6.3}{63} = 0.10\text{ eq}$$

This is contained in $$250\text{ mL} = 0.25\text{ L}$$, so the normality $$N_1$$ of the given oxalic-acid solution is

$$N_1 = \frac{0.10\ \text{eq}}{0.25\ \text{L}} = 0.40\text{ N}$$

Let $$V_2$$ be the volume (in mL) of $$0.1\text{ N}$$ NaOH required to neutralize $$V_1 = 10\text{ mL}$$ of the acid. At equivalence, the number of gram-equivalents of acid equals that of base:

$$N_1 V_1 = N_2 V_2$$

$$0.40 \times 10 = 0.10 \times V_2$$

$$V_2 = \frac{0.40 \times 10}{0.10} = 40\text{ mL}$$

Therefore, $$40\text{ mL}$$ of $$0.1\text{ N}$$ NaOH is required.

Option D which is: $$40$$ ml