Here water is solvent and ethylene glycol is solute.

We know, Depression of freezing point,

$$\Delta $$T_f = K_f . m

$$\Delta {T_f} = 1.86 \times {{\left( {{{62} \over {62}}} \right)} \over {\left( {{{250} \over {1000}}} \right)}}$$

= 7.44

We know, freezing point of pure water (here solvent) is 0^oC. Which is represented by $$T_f^0$$.

We also know, $$\Delta {T_f} = T_f^0 - {T_f}$$

where $$T_f^0$$ = Freezing point of pure solvent and $${T_f}$$ = Freezing point of solution

$$ \therefore $$ 7.44 = 0 - $${T_f}$$

$$ \Rightarrow $$ $${T_f}$$ = -7.44^oC

So, not a single drop of water solution will become ice until temperature reaches -7.44^oC. When temperature decrease more than -7.44^oC then some part of the solution starts becoming ice staring from surface of the solution.

Now let W_l gm of solution still stays in liquid phase when temperature reaches -10^oC.

$$ \therefore $$ $$10 = 1.86 \times {{\left( {{{62} \over {62}}} \right)} \over {\left( {{{{W_l}} \over {1000}}} \right)}}$$

W_l = 186 gm

So, The amount of water (in g) separated as ice is

$$\Delta $$W = (250 $$-$$ 186) = 64 gm