If one can find a student with at least $k$ friends in any class which has $21$ students such that at least two of  any three of these students are friends, what is the largest possible value of $k$?

$ 
\textbf{(A)}\ 8
\qquad\textbf{(B)}\ 9
\qquad\textbf{(C)}\ 10
\qquad\textbf{(D)}\ 11
\qquad\textbf{(E)}\ 12
$
1. Let \( A \) be the student with the most friends, and let \( B \) be a student who is not a friend of \( A \). Let \( X \) be one of the other 19 students in the class.

2. For the triplet \( A, B, X \), since at least two of any three students are friends, \( X \) must be a friend of either \( A \) or \( B \).

3. Define \( X(A) \) as the number of friends of \( A \) and \( X(B) \) as the number of friends of \( B \). By the principle of inclusion and exclusion, we have:
   \[
   X(A) + X(B) - X(A \cap B) = 19
   \]
   where \( X(A \cap B) \) is the number of common friends between \( A \) and \( B \).

4. Since \( X(A) + X(B) \geq 19 \) and \( X(A) \geq X(B) \), it follows that:
   \[
   2 \cdot X(A) \geq 19
   \]
   Therefore,
   \[
   X(A) \geq \frac{19}{2} = 9.5
   \]

5. Since \( X(A) \) must be an integer, the minimal value of \( X(A) \) is 10.

6. Thus, in any class of 21 students, there exists a student \( A \) with at least 10 friends.

Conclusion:
\[
\boxed{10}
\]