The molarity of ${\mathrm{H}}_3{\mathrm{PO}}_4(\mathrm{d}=1.8\mathrm{g}/\mathrm{mL})$ is 18 M. Hence, mass percentage and molality are: 

Density = 1.8 g/mL
Molarity = 18 M = 18 mol H₃PO₄ in 1 L solution
1000 mL solution has H₃PO₄= 18 mol = 18 × 98 g
100 mL = (1800 g) solution has H₃PO₄ = 18 × 98 g
H₃PO₄ by mass o/o (by weight of solution) 

$=\frac{18\times 98}{1800}\times 100=98\mathrm{\%}$

or $(1800-18\times 98){\mathrm{gH}}_2\mathrm{O}\text{ (solvent) }{\mathrm{H}}_3{\mathrm{PO}}_4=18\times 98\mathrm{g}$

$2\mathrm{g}\text{ solvent has }{\mathrm{H}}_3{\mathrm{PO}}_4=98\mathrm{g}=1{\mathrm{molH}}_3{\mathrm{PO}}_4$

$1\mathrm{g}=1000\mathrm{g}\text{ solvent has }{\mathrm{H}}_3{\mathrm{PO}}_4=500\text{ molal }$

Thus, 98%, 500 molal.