［Solution 1］From the given information, Alice walks from $A$ to $K$ (as shown in the figure).
It is easy to calculate that $A C=2 \sqrt{3}$, and $\angle A C K=90^{\circ}$
$\left(\because \angle B C D=120^{\circ}\right.$, and $\left.\angle B C A=30^{\circ}\right)$.
$$
\begin{aligned}
\therefore \quad A K & =\sqrt{A C^{2}+C K^{2}} \\
& =\sqrt{12+1}=\sqrt{13} .
\end{aligned}
$$

Therefore, the answer is $(A)$.
[Solution 2] Connect $A D$, it is easy to find that $A D=4$, and $\angle A D K=60^{\circ}, D K=1$. In $\triangle A D K$, by the cosine rule we have
$$
\begin{aligned}
A K & =\sqrt{A D^{2}+D K^{2}-2 A D \cdot D K \cos 60^{\circ}} \\
& =\sqrt{4^{2}+1^{2}-2 \cdot 4 \cdot 1 \cdot \frac{1}{2}} \\
& =\sqrt{13} .
\end{aligned}
$$

Therefore, the answer is $(A)$.
19.6 A park is in the shape of a regular hexagon with each side 2 kilometers long. Alice starts from a corner and walks 5 kilometers along the perimeter of the park. The distance in kilometers from her starting point is 
(A) $\sqrt{13}$.
(B) $\sqrt{14}$.
(C) $\sqrt{15}$.
(D) $\sqrt{16}$.
(E) $\sqrt{17}$.
(37th American High School Mathematics Examination, 1986)