At summer camp, there are $20$ campers in each of the swimming class, the archery class, and the rock climbing class. Each camper is in at least one of these classes. If $4$ campers are in all three classes, and $24$ campers are in exactly one of the classes, how many campers are in exactly two classes?

$\text{(A) }10\qquad\text{(B) }11\qquad\text{(C) }12\qquad\text{(D) }13\qquad\text{(E) }14$
1. Let \( S \), \( A \), and \( R \) represent the sets of campers in the swimming, archery, and rock climbing classes, respectively. We are given:
   \[
   |S| = |A| = |R| = 20
   \]
   Each camper is in at least one of these classes.

2. Let \( n \) be the total number of campers. We know:
   \[
   n = |S \cup A \cup R|
   \]

3. We are given:
   \[
   |S \cap A \cap R| = 4
   \]
   and
   \[
   24 \text{ campers are in exactly one class}
   \]

4. Let \( x \) be the number of campers in exactly two classes. We need to find \( x \).

5. Using the principle of inclusion-exclusion for three sets, we have:
   \[
   |S \cup A \cup R| = |S| + |A| + |R| - |S \cap A| - |A \cap R| - |R \cap S| + |S \cap A \cap R|
   \]

6. Substituting the known values:
   \[
   n = 20 + 20 + 20 - |S \cap A| - |A \cap R| - |R \cap S| + 4
   \]
   Simplifying:
   \[
   n = 60 - (|S \cap A| + |A \cap R| + |R \cap S|) + 4
   \]
   \[
   n = 64 - (|S \cap A| + |A \cap R| + |R \cap S|)
   \]

7. Let \( |S \cap A| + |A \cap R| + |R \cap S| = y \). Then:
   \[
   n = 64 - y
   \]

8. We know that 24 campers are in exactly one class. Let \( a \), \( b \), and \( c \) be the number of campers in exactly one of the swimming, archery, and rock climbing classes, respectively. Then:
   \[
   a + b + c = 24
   \]

9. The total number of campers counted in the three classes is:
   \[
   20 + 20 + 20 = 60
   \]

10. The 4 campers in all three classes are counted three times, so they contribute:
    \[
    3 \times 4 = 12
    \]

11. The \( x \) campers in exactly two classes are counted twice, so they contribute:
    \[
    2x
    \]

12. The 24 campers in exactly one class are counted once, so they contribute:
    \[
    24
    \]

13. Therefore, the equation for the total count is:
    \[
    24 + 2x + 12 = 60
    \]

14. Solving for \( x \):
    \[
    24 + 2x + 12 = 60
    \]
    \[
    36 + 2x = 60
    \]
    \[
    2x = 24
    \]
    \[
    x = 12
    \]

The final answer is \( \boxed{12} \)