First, find the molarity of the concentrated sulphuric acid solution. Molarity ($M_1$) = $\frac{\text{Mass percentage} \times 10 \times \text{Density}}{\text{Molar mass}}$ Molar mass of $\text{H}_2\text{SO}_4 = (2 \times 1) + 32 + (4 \times 16) = 98\text{ g mol}^{-1}$. $M_1 = \frac{98 \times 10 \times 1.80}{98} = 18\text{ M}$. Now, use the dilution formula to find the volume required ($V_1$) to prepare $1\text{ L}$ ($1000\text{ mL}$) of $0.1\text{ M}$ $\text{H}_2\text{SO}_4$ solution. $M_1V_1 = M_2V_2$ $18 \times V_1 = 0.1 \times 1000$ $18 \times V_1 = 100$ $V_1 = \frac{100}{18} \approx 5.55\text{ mL}$.