1. Given data

   • Molarity of solution: $0.8\,M$
   • Density of solution: $1.06\,\text{g cm}^{-3} = 1.06\,\text{g mL}^{-1}$
   • Solute: $\mathrm{H_2SO_4}$

2. Take 1 litre of solution Since molarity is given, in $1\,\text{L}$ solution: $\text{moles of } \mathrm{H_2SO_4} = 0.8$

3. Mass of 1 litre solution $\text{mass of solution} = 1000\,\text{mL} \times 1.06\,\frac{\text{g}}{\text{mL}} = 1060\,\text{g}$

4. Mass of solute Molar mass of $\mathrm{H_2SO_4}$: $2(1) + 32 + 4(16) = 98\,\text{g mol}^{-1}$ So, mass of $0.8$ mol acid: $0.8 \times 98 = 78.4\,\text{g}$

5. Mass of solvent $\text{mass of solvent} = 1060 - 78.4 = 981.6\,\text{g} = 0.9816\,\text{kg}$

6. Calculate molality $m = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{0.8}{0.9816}$ $m \approx 0.81499\,\text{mol kg}^{-1}$ $m \approx 0.815\,m$

7. Match with asked format The question writes molality as: $\underline{\hspace{2cm}}\times 10^{-3}\,m$ But the calculated molality is clearly: $0.815\,m = 815 \times 10^{-3}\,m$

Therefore, the required integer is: $\boxed{815}$