Similar Triangles!

Since $\triangle ABC$ and $\triangle DEF$ are similar to each other, $\angle B = \angle F$ , and since the ratio of $AB$ to $EF$ is $1 : 2$ , $EF = 2AB$ , and since the ratio of $BC$ to $DE$ is $1 : 3$ , $DE = 3BC$ .

The area of $\triangle ABC$ is $\frac{1}{2} \cdot AB \cdot BC \sin \angle B$ and the area of $\triangle DEF$ is $\frac{1}{2} \cdot DE \cdot EF \sin \angle E = \frac{1}{2} \cdot (3BC) \cdot (2AB) \sin \angle B$ .

The ratio of the areas of the two triangles is therefore $\frac{\frac{1}{2} \cdot AB \cdot BC \sin \angle B}{\frac{1}{2} \cdot (3BC) \cdot (2AB) \sin \angle B} = \boxed{\frac{1}{6}}$ .