1. Understand molality

A $100$ molal aqueous solution means:

$m = 100 = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$

So, for convenience, take $1$ kg of water as solvent.

Then,

$\text{moles of solute} = 100$

2. Find moles of water

Molar mass of water:

$M_{\mathrm{H_2O}} = 2(1) + 16 = 18\ \text{g mol}^{-1}$

Mass of water taken:

$1\ \text{kg} = 1000\ \text{g}$

Hence moles of water:

$n_{\mathrm{H_2O}} = \frac{1000}{18} = 55.56$

3. Calculate mole fraction of solute

Let mole fraction of solute be $x$.

$x = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\mathrm{H_2O}}}$

Substitute values:

$x = \frac{100}{100 + 55.56}$

$x = \frac{100}{155.56} \approx 0.6429$

4. Match with required format

We need to write mole fraction as:

$\underline{\hspace{2cm}} \times 10^{-2}$

Now,

$0.6429 = 64.29 \times 10^{-2}$

Rounded to nearest integer:

$64$

5. Comparison with stored answer

Derived answer = $64$

Stored correct answer = $64$

They match.