Use the graph information

The diagram (freezing curve) indicates that after adding methanol, the freezing point of water is lowered by $\Delta T_f = 2.5\,\text{K}$ (i.e. from $273.15\,\text{K}$ to $270.65\,\text{K}$).

Apply freezing point depression formula

For a non-electrolyte like methanol, $\Delta T_f = K_f m$ where $m$ is molality.

So, $m = \frac{\Delta T_f}{K_f} = \frac{2.5}{1.86} \approx 1.344\,\text{mol kg}^{-1}$

Mass of solvent (water)

Volume of water $=100\,\text{mL}$ and density $=1\,\text{g mL}^{-1}$ $\Rightarrow \text{mass of water} = 100\,\text{g} = 0.100\,\text{kg}$

Moles of methanol required

Using $m = \frac{n_{\text{solute}}}{\text{kg of solvent}}$ we get $n_{\text{methanol}} = m \times 0.100 = 1.344 \times 0.100 = 0.1344\,\text{mol}$

Mass of methanol

Molar mass of methanol $=32\,\text{g mol}^{-1}$ $\text{mass} = 0.1344 \times 32 \approx 4.30\,\text{g}$

Convert mass to volume

Density of methanol $=0.792\,\text{g mL}^{-1}$ $V = \frac{4.30}{0.792} \approx 5.43\,\text{mL}$

Find $x$

Given volume added is $x \times 10^{-2}\,\text{mL}$, $x \times 10^{-2} = 5.43$ $x = 5.43 \times 10^2 = 543$

Final answer

$\boxed{x=543}$