1. Molar mass of acetic acid

Acetic acid is $\mathrm{CH_3COOH} = \mathrm{C_2H_4O_2}$.

$M = 2(12) + 4(1) + 2(16) = 24 + 4 + 32 = 60\ \text{g mol}^{-1}$

2. Moles of acetic acid added

Given mass $= 5\ \text{g}$,

$n = \frac{5}{60} = \frac{1}{12}\ \text{mol}$

3. Mass of solvent (water)

Volume of water $= 500\ \text{mL}$ and density $= 1\ \text{g mL}^{-1}$,

so mass of water

$= 500\ \text{g} = 0.5\ \text{kg}$

4. Molality before dissociation correction

$m = \frac{1/12}{0.5} = \frac{1}{6}\ \text{mol kg}^{-1}$

5. van't Hoff factor

Acetic acid dissociates as

$\mathrm{CH_3COOH \rightleftharpoons CH_3COO^- + H^+}$

If degree of dissociation is $\alpha = 20\% = 0.2$, then for a solute giving 2 particles,

$i = 1 + \alpha = 1 + 0.2 = 1.2$

6. Depression in freezing point

$\Delta T_f = i K_f m$

$\Delta T_f = (1.2)(1.86)\left(\frac{1}{6}\right)$

$\Delta T_f = 1.2 \times 0.31 = 0.372\ ^\circ\mathrm{C}$

7. Write in the required form

$0.372\ ^\circ\mathrm{C} = 372 \times 10^{-3}\ ^\circ\mathrm{C}$

Therefore, the required integer is:

$\boxed{372}$