A square flag has a red cross of uniform width with a blue square in the center on a white background as shown. (The cross is symmetric with respect to each of the diagonals of the square.) If the entire cross (both the red arms and the blue center) takes up $36\%$ of the area of the flag, what percent of the area of the flag is blue?
[asy]
draw((0,0)--(5,0)--(5,5)--(0,5)--cycle);
draw((1,0)--(5,4));
draw((0,1)--(4,5));
draw((0,4)--(4,0));
draw((1,5)--(5,1));
label("RED", (1.2,3.7));
label("RED", (3.8,3.7));
label("RED", (1.2,1.3));
label("RED", (3.8,1.3));
label("BLUE", (2.5,2.5));
[/asy] 
$ \textbf{(A)}\ 0.5 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 6 $
1. Let the area of the entire square flag be \(100\) square units for simplicity. This means the area of the cross (both the red arms and the blue center) is \(36\%\) of \(100\), which is \(36\) square units.
2. The remaining area, which consists of the four white triangles, is \(100 - 36 = 64\) square units. Each of these four triangles has an area of \(\frac{64}{4} = 16\) square units.
3. Let the side length of the square flag be \(10\) units. The side length of the blue square in the center is denoted as \(s\).
4. The width of each arm of the cross is \(\sqrt{2}s\), since the cross is symmetric with respect to the diagonals of the square.
5. The diagonal of the blue square is \(2s\), and the side length of the blue square is \(s\).
6. The area of the blue square is \(s^2\).
7. The area of the red cross (excluding the blue square) is \(36 - s^2\).
8. The area of each white triangle is \(\frac{1}{2} \times (10 - 2s) \times (10 - 2s) = 16\).
9. Solving for \(s\), we have:
   \[
   (10 - 2s)^2 = 16 \implies 100 - 40s + 4s^2 = 16 \implies 4s^2 - 40s + 84 = 0 \implies s^2 - 10s + 21 = 0
   \]
10. Solving the quadratic equation \(s^2 - 10s + 21 = 0\) using the quadratic formula \(s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
    \[
    s = \frac{10 \pm \sqrt{100 - 84}}{2} = \frac{10 \pm \sqrt{16}}{2} = \frac{10 \pm 4}{2}
    \]
    \[
    s = 7 \quad \text{or} \quad s = 3
    \]
11. Since \(s = 7\) is not possible (as it would make the side length of the blue square larger than the side length of the flag), we have \(s = 3\).
12. The area of the blue square is \(s^2 = 3^2 = 9\) square units.

The final answer is \(\boxed{2}\)