In how many ways can $7$ red, $7$ white balls be distributed into $7$ boxes such that every box contains exactly $2$ balls?

$ \textbf{(A)}\ 163
\qquad\textbf{(B)}\ 393
\qquad\textbf{(C)}\ 858
\qquad\textbf{(D)}\ 1716
\qquad\textbf{(E)}\ \text{None}
$
To solve this problem, we need to distribute 7 red and 7 white balls into 7 boxes such that each box contains exactly 2 balls. We will consider different cases based on the number of boxes that contain 2 white balls.

1. **Case 1: 3 boxes with 2 white balls**
   - In this case, we will have 3 boxes with 2 white balls, 1 box with 1 white ball, and 3 boxes with 0 white balls.
   - The number of ways to choose which 3 boxes out of 7 will contain 2 white balls is given by the multinomial coefficient:
     \[
     \frac{7!}{3!1!3!} = \frac{5040}{6 \cdot 1 \cdot 6} = 140
     \]

2. **Case 2: 2 boxes with 2 white balls**
   - In this case, we will have 2 boxes with 2 white balls, 3 boxes with 1 white ball, and 2 boxes with 0 white balls.
   - The number of ways to choose which 2 boxes out of 7 will contain 2 white balls is given by the multinomial coefficient:
     \[
     \frac{7!}{2!3!2!} = \frac{5040}{2 \cdot 6 \cdot 2} = 210
     \]

3. **Case 3: 1 box with 2 white balls**
   - In this case, we will have 1 box with 2 white balls, 5 boxes with 1 white ball, and 1 box with 0 white balls.
   - The number of ways to choose which 1 box out of 7 will contain 2 white balls is given by the multinomial coefficient:
     \[
     \frac{7!}{1!5!1!} = \frac{5040}{1 \cdot 120 \cdot 1} = 42
     \]

4. **Case 4: 0 boxes with 2 white balls**
   - In this case, each box will have one red ball and one white ball.
   - There is only 1 way to distribute the balls in this manner:
     \[
     1
     \]

Adding up all the cases, we get the total number of ways to distribute the balls:
\[
140 + 210 + 42 + 1 = 393
\]

The final answer is $\boxed{393}$.