Each side of triangle $ABC$ is $12$ units. $D$ is the foot of the perpendicular dropped from $A$ on $BC$, and $E$ is the midpoint of $AD$. The length of $BE$, in the same unit, is:

${{ \textbf{(A)}\ \sqrt{18} \qquad\textbf{(B)}\ \sqrt{28} \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ \sqrt{63} }\qquad\textbf{(E)}\ \sqrt{98} } $
1. **Identify the properties of the equilateral triangle:**
   Since $\triangle ABC$ is equilateral with each side equal to $12$ units, all angles are $60^\circ$. The altitude from $A$ to $BC$ will also be the median and the angle bisector.

2. **Calculate the length of the altitude $AD$:**
   In an equilateral triangle, the altitude can be calculated using the formula for the height of an equilateral triangle:
   \[
   h = \frac{\sqrt{3}}{2} \times \text{side length}
   \]
   Therefore,
   \[
   AD = \frac{\sqrt{3}}{2} \times 12 = 6\sqrt{3}
   \]

3. **Determine the lengths of $BD$ and $CD$:**
   Since $D$ is the foot of the perpendicular from $A$ to $BC$ and $ABC$ is equilateral, $D$ is the midpoint of $BC$. Thus,
   \[
   BD = DC = \frac{BC}{2} = \frac{12}{2} = 6
   \]

4. **Find the midpoint $E$ of $AD$:**
   Since $E$ is the midpoint of $AD$,
   \[
   AE = \frac{AD}{2} = \frac{6\sqrt{3}}{2} = 3\sqrt{3}
   \]

5. **Apply the Pythagorean Theorem in $\triangle BED$:**
   $\triangle BED$ is a right triangle with $BD = 6$ and $ED = AE = 3\sqrt{3}$. Using the Pythagorean Theorem:
   \[
   BE^2 = BD^2 + ED^2
   \]
   Substituting the known values:
   \[
   BE^2 = 6^2 + (3\sqrt{3})^2 = 36 + 27 = 63
   \]
   Therefore,
   \[
   BE = \sqrt{63}
   \]

Conclusion:
\[
BE = \boxed{\sqrt{63}}
\]