Solution:

The answer is $\mathbf{( C )}$. Since the number is a palindrome, we can write it as $a b c c b a$, where $a, b, c$ are digits and $a \neq 0$. The divisibility criterion for $11$ is then automatically satisfied, since $a-b+c-c+b-a=0$.

For the divisibility criterion for $3$ it is necessary that $2(a+b+c)$ be a multiple of $3$, that is, that the number $a b c$ be a multiple of $3$; the answer is therefore the number of integers $100 \leq n \leq 999$ that are multiples of $3$, which are $\frac{999-102}{3}+1=300$.