$1000$ ml solution contains $0.8$mol of H₂SO₄

Mass of solution = $100{\mathrm{cm}}^3\times 1.06\frac{ \mathrm{g}}{{\mathrm{cm}}^3}= 1060 \mathrm{g}$

Mass of solute (H₂SO₄) =$0.8 \times 98 =78.4\mathrm{g}$

Mass of solvent = $1060 -78.4 =981.6 \mathrm{g}$

$\mathrm{Molality} = \frac{\mathrm{moles} \mathrm{of} {\mathrm{H}}_2{\mathrm{SO}}_4}{\mathrm{mass} \mathrm{of} \mathrm{solvent} (\mathrm{in} \mathrm{kg})}$

$=\frac{0.8}{981.6\times {10}^{-3}} = 0.815\mathrm{m}$

= $815\times {10}^{-3} \mathrm{m}$