The area of the smallest square that will contain a circle of radius 4 is

$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 64 \qquad \textbf{(E)}\ 128$
1. To find the area of the smallest square that can contain a circle of radius 4, we first need to determine the side length of the square.
2. The diameter of the circle is twice the radius. Given the radius \( r = 4 \), the diameter \( d \) is:
   \[
   d = 2r = 2 \times 4 = 8
   \]
3. The smallest square that can contain the circle must have a side length equal to the diameter of the circle. Therefore, the side length of the square is 8.
4. The area \( A \) of a square is given by the formula:
   \[
   A = \text{side length}^2
   \]
5. Substituting the side length of 8 into the formula, we get:
   \[
   A = 8^2 = 64
   \]

The final answer is \(\boxed{64}\)