We know,

$$\Delta $$T_f = i $$ \times $$ k_f $$ \times $$ m

$$ \therefore $$ $$\Delta $$T_{fDil. Milk} = (1) $$ \times $$ k_f $$ \times $$ m_dil = 0.2 ...(1)

$$\Delta $$T_{fPure Milk} = (1) $$ \times $$ k_f $$ \times $$ m_pure = 0.5 ...(2)

From (1) and (2), we get

$${{{m_{pure}}} \over {{m_{dil}}}} = {{0.5} \over {0.2}}$$ = $${5 \over 2}$$ ....(3)

Assume moles of fat in both the milk samples = n

Weight of milk = w₁

and weight of water = w₀

Molality of pure milk,
m_pure = $${n \over {{w_1}}} \times 1000$$ ....(4)

Molality of diluted milk,
m_dil. = $${n \over {{w_1} + {w_0}}} \times 1000$$ ....(5)

From (4) and (5), we get

$${{{m_{pure}}} \over {{m_{dil}}}} = {{{w_1} + {w_0}} \over {{w_1}}}$$

Using equation (3), we get

$${5 \over 2} = {{{w_1} + {w_0}} \over {{w_1}}}$$

$$ \Rightarrow $$ 5w₁ = 2w₁ + 2w₀

$$ \Rightarrow $$ 3w₁ = 2w₀

$$ \Rightarrow $$ $${{{w_1}} \over {{w_0}}} = {2 \over 3}$$

Which means 3 cups of water has been added to 2 cups of pure milk.