Many television screens are rectangles that are measured by the length of their diagonals. The ratio of the horizontal length to the height in a standard television screen is $ 4 : 3$. The horizontal length of a “$ 27$-inch” television screen is closest, in inches, to which of the following?
[asy]import math;
unitsize(7mm);
defaultpen(linewidth(.8pt)+fontsize(8pt));

draw((0,0)--(4,0)--(4,3)--(0,3)--(0,0)--(4,3));
fill((0,0)--(4,0)--(4,3)--cycle,mediumgray);
label(rotate(aTan(3.0/4.0))*"Diagonal",(2,1.5),NW);
label(rotate(90)*"Height",(4,1.5),E);
label("Length",(2,0),S);[/asy]$ \textbf{(A)}\ 20 \qquad
\textbf{(B)}\ 20.5 \qquad
\textbf{(C)}\ 21 \qquad
\textbf{(D)}\ 21.5 \qquad
\textbf{(E)}\ 22$
1. Let the horizontal length of the television screen be \( a \) and the height be \( b \). According to the problem, the ratio of the horizontal length to the height is \( 4:3 \). Therefore, we can write:
   \[
   \frac{a}{b} = \frac{4}{3} \implies a = \frac{4}{3}b
   \]

2. The diagonal of the television screen is given as 27 inches. Using the Pythagorean theorem for the right triangle formed by the horizontal length, height, and diagonal, we have:
   \[
   a^2 + b^2 = 27^2
   \]

3. Substitute \( a = \frac{4}{3}b \) into the Pythagorean theorem equation:
   \[
   \left(\frac{4}{3}b\right)^2 + b^2 = 27^2
   \]

4. Simplify the equation:
   \[
   \left(\frac{4}{3}b\right)^2 = \frac{16}{9}b^2
   \]
   \[
   \frac{16}{9}b^2 + b^2 = 27^2
   \]
   \[
   \frac{16}{9}b^2 + \frac{9}{9}b^2 = 729
   \]
   \[
   \frac{25}{9}b^2 = 729
   \]

5. Solve for \( b^2 \):
   \[
   b^2 = 729 \times \frac{9}{25}
   \]
   \[
   b^2 = 262.44
   \]
   \[
   b = \sqrt{262.44} \approx 16.2
   \]

6. Now, find \( a \) using \( a = \frac{4}{3}b \):
   \[
   a = \frac{4}{3} \times 16.2 \approx 21.6
   \]

7. The horizontal length \( a \) is approximately 21.6 inches. The closest value to 21.6 inches among the given choices is 21.5 inches.

The final answer is \( \boxed{21.5} \)