Given,
$$
\begin{aligned}
& \text { Mass of } \mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]=0 \mathrm{lg} \\
& \text { Mass of solvent }=100 \mathrm{~g} \\
& \text { Molar mass of } \mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]=329 \mathrm{~g} \mathrm{~mol}^{-1} \\
& k_f=1.86 \mathrm{~K} \mathrm{Kg} / \mathrm{mol} \\
& \mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right] \rightleftharpoons \\
& \left.i \mathrm{~K}^{+}+\mathrm{Fe}(\mathrm{CN})_6\right]^{-} \\
& i\left(\operatorname{van}^{\prime} \mathrm{t} \text { Hoff factor }\right)=3+1=4
\end{aligned}
$$
From, depression in freezing point,
$$
\begin{aligned}
& T_f=i \times K_f \times m \\
& T_f=4 \times 1.86 \mathrm{Kkg} / \mathrm{moI} \times \frac{0 \mathrm{gg}}{329 \mathrm{~g}} \times \frac{1000}{100 \mathrm{~kg}} \\
& \mathrm{~T}_f=0.0226 \mathrm{~K}
\end{aligned}
$$
Thus, option (3) is correct.