B, neutralizes 25cm3 of 0.05M sulphuric acid solution?

Sulfuric acid ($\text{H}_2\text{SO}_4$) reacts with sodium hydroxide ($\text{NaOH}$) to produce sodium sulfate ($\text{Na}_2\text{SO}_4$) and water ($\text{H}_2\text{O}$).

$$\text{H}_2\text{SO}_4\text{(aq)} + 2\text{NaOH}\text{(aq)} \longrightarrow \text{Na}_2\text{SO}_4\text{(aq)} + 2\text{H}_2\text{O}\text{(l)}$$

From the balanced equation, 1 mole of $\text{H}_2\text{SO}_4$ reacts with 2 moles of $\text{NaOH}$.

Given:
Volume of $\text{H}_2\text{SO}_4$ solution = 25 $\text{cm}^3$ = 0.025 L
Concentration of $\text{H}_2\text{SO}_4$ solution = 0.05 M

$$\text{Moles of H}_2\text{SO}_4 = \text{Molarity} \times \text{Volume (L)}$$ $$\text{Moles of H}_2\text{SO}_4 = 0.05 \text{ mol/L} \times 0.025 \text{ L}$$ $$\text{Moles of H}_2\text{SO}_4 = 0.00125 \text{ mol}$$

Using the mole ratio from the balanced equation (1 mole $\text{H}_2\text{SO}_4$ : 2 moles $\text{NaOH}$):

$$\text{Moles of NaOH} = \text{Moles of H}_2\text{SO}_4 \times \frac{2 \text{ mol NaOH}}{1 \text{ mol H}_2\text{SO}_4}$$ $$\text{Moles of NaOH} = 0.00125 \text{ mol} \times 2$$ $$\text{Moles of NaOH} = 0.0025 \text{ mol}$$

Given:
Concentration of $\text{NaOH}$ solution = 0.1 M

$$\text{Volume of NaOH (L)} = \frac{\text{Moles of NaOH}}{\text{Molarity of NaOH}}$$ $$\text{Volume of NaOH (L)} = \frac{0.0025 \text{ mol}}{0.1 \text{ mol/L}}$$ $$\text{Volume of NaOH (L)} = 0.025 \text{ L}$$

$$\text{Volume of NaOH (cm}^3) = 0.025 \text{ L} \times \frac{1000 \text{ cm}^3}{1 \text{ L}}$$ $$\text{Volume of NaOH (cm}^3) = 25 \text{ cm}^3$$

The volume of 0.1 M sodium hydroxide solution required is $\boxed{\text{0.025 L}}$ or $\boxed{\text{25 cm}^3}$.