Correct Answer: $21:4$

Solution :

Given that in $\triangle ABC$, $D$ and $E$ are the points of sides $AB$ and $BC$, $DE \parallel AC$ and $AD : DB = 3 : 2$.
From the figure,
$\frac{DB}{AB}=\frac{2}{5}$
In $\triangle ABC$ and $\triangle DBE$,
$\angle ABC=\angle DBE$ (common angle)
$\angle BAC=\angle BDE$ (corresponding angle)
$\angle ACB=\angle DEB$ (corresponding angle)
$\triangle ABC\sim\triangle DBE$, by AAA similarity.
By Thales theorem, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\therefore\frac{\text{area of $\triangle ABC$}}{\text{area of $\triangle DBE$}}=(\frac{AB}{DB})^2$
$⇒\frac{\text{area of $\triangle ABC$}}{\text{area of $\triangle DBE$}}=(\frac{5}{2})^2$
$⇒\frac{\text{area of $\triangle ABC$}}{\text{area of $\triangle DBE$}}=\frac{25}{4}$
$⇒\frac{\text{area of $\triangle ABC$}}{\text{area of $\triangle DBE$}}-1 =\frac{25}{4}-1$
$⇒\frac{\text{area of $\triangle ABC$}-\text{area of $\triangle DBE$}}{\text{area of $\triangle DBE$}} =\frac{25-4}{4}$
$⇒\frac{\text{area of trapezium $ACED$ }}{\text{area of $\triangle DBE$}} =\frac{21}{4}$
Hence, the correct answer is $21:4$.