ABC and BDE are two equilateral triangles such that D is the mid point of BC.

Given: ABC and BDE are two equilateral triangles
Since, D is the mid point of BC and BDE is also an equilateral traingle.
Hence, E is also the mid point of AB.
Now, D and E are the mid points of BC and AB.
In a triangle, the line segment that joins the midpoints of the two sides of a triangle is parallel to the third side and is half of it.

$\mathrm{DE} \| \mathrm{CA}$ and $\mathrm{DE}=\frac{1}{2} \mathrm{CA}$

Now, In △ABC and △EBD
∠BED = ∠BAC             (Corresponding angles)
∠B = ∠B             (Common)
By AA-similarity criterion
△ABC ∼ △EBD
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.

$\therefore \frac{\operatorname{area}(\triangle \mathrm{ABC})}{\operatorname{area}(\triangle \mathrm{DBE})}=\left(\frac{\mathrm{AC}}{\mathrm{ED}}\right)^{2}=\left(\frac{2 \mathrm{ED}}{\mathrm{ED}}\right)^{2}=\frac{4}{1}$

Hence, the correct answer is option (d).