A rhombus is formed by two radii and two chords of a circle whose radius is $ 16$ feet. The area of the rhombus in square feet is:

$ \textbf{(A)}\ 128 \qquad\textbf{(B)}\ 128\sqrt {3} \qquad\textbf{(C)}\ 256 \qquad\textbf{(D)}\ 512 \qquad\textbf{(E)}\ 512\sqrt {3}$
1. **Identify the shorter diagonal:**
   The shorter diagonal of the rhombus is one of the radii of the circle, which is given as \(16\) feet.

2. **Determine the properties of the rhombus:**
   Since the rhombus is formed by two radii and two chords of the circle, the shorter diagonal divides the rhombus into two congruent isosceles triangles. Each of these triangles has two sides that are radii of the circle, making them equilateral triangles.

3. **Calculate the longer diagonal:**
   In an equilateral triangle, the altitude (which is half of the longer diagonal of the rhombus) can be found using the Pythagorean theorem. For an equilateral triangle with side length \(16\) feet, the altitude \(h\) is given by:
   \[
   h = \sqrt{16^2 - 8^2} = \sqrt{256 - 64} = \sqrt{192} = 8\sqrt{3}
   \]
   Therefore, the longer diagonal of the rhombus is twice the altitude:
   \[
   \text{Longer diagonal} = 2 \times 8\sqrt{3} = 16\sqrt{3}
   \]

4. **Calculate the area of the rhombus:**
   The area \(A\) of a rhombus can be calculated using the formula involving its diagonals \(d_1\) and \(d_2\):
   \[
   A = \frac{1}{2} \times d_1 \times d_2
   \]
   Here, \(d_1 = 16\) feet and \(d_2 = 16\sqrt{3}\) feet. Substituting these values in, we get:
   \[
   A = \frac{1}{2} \times 16 \times 16\sqrt{3} = \frac{1}{2} \times 256\sqrt{3} = 128\sqrt{3}
   \]

5. **Conclusion:**
   The area of the rhombus is \(128\sqrt{3}\) square feet.

The final answer is \(\boxed{128\sqrt{3}}\).