$100$ molal aqueous solution means there is $100$ mole solute in $1 \mathrm{kg}=1000 \mathrm{gm}$ water.

Now,

mole-fraction of solute$=\frac{{\mathrm{n}}_{\text{solute }}}{{\mathrm{n}}_{\text{solute }}+{\mathrm{n}}_{\text{solvent }}}$

$=\frac{100}{100+\frac{1000}{18}}=\frac{1800}{2800}=0.6428$

$=64.28\times {10}^{-2}$