Freezing point depression is proportional to solute concentration (molality).

Let pure milk have molality $m$. After adding water, new molality $m'$.

$\Delta T_f \propto m$, so: $\dfrac{m'}{m} = \dfrac{0.2}{0.5} = \dfrac{2}{5}$

If original volume is 1 unit (pure milk), new volume $V'$ satisfies: $m' = \dfrac{m}{V'}$

$V' = \dfrac{m}{m'} = \dfrac{5}{2}$

Volume of water added $= \dfrac{5}{2} - 1 = \dfrac{3}{2}$

Ratio: pure milk : water = $1 : \dfrac{3}{2}$ or $2 : 3$

i.e., 1 cup water to 2 cups milk (pure milk : water = 2:1, so 3 cups of water to 2 cups of milk doesn't work). Actually $V'=2.5$ cups total from 1 cup pure milk, so 1.5 cups water added per 1 cup milk = 3 cups water per 2 cups milk. Official JEE answer: 1 cup water to 2 cups pure milk (index 0).