Since $2010$ is neither a prime nor a palindromic number, $k > 1$. Suppose $k = 2$. Then there are two palindromic primes $a, b$ such that $2010 = a + b$. Since $1005$ isn't prime, $a \neq b$. Say $a < b$. Then $2 \le a < 1005 < b \le 2009$. But palindromic numbers $b > 2$ ending in $2$ are even, and so can't be prime. Hence, $b \le 1991$. Also, every number of the form $1ax1$, where $x \in \{0, 1, \dots, 9\}$, is divisible by $11$. Hence, there are no palindromic primes of this form, and so $b < 999$. This means that $2010 = a + b < 2b < 2000$, which is absurd. Hence, $k \ge 3$.

If $k = 3$, then there are three palindromic primes $a, b, c$ such that $2010 = a + b + c$. If one of $a, b, c$ is even, say $a = 2$, then $2008 = b + c$ and the argument just given shows that this can't happen. But the sum of three odd numbers is always odd, hence never equal to $2010$. Thus, $k \ge 4$.

In fact, $k = 4$, since
$$2010 = 11 + 929 + 919 + 151 = 11 + 929 + 757 + 313 = 11 + 919 + 727 + 353$$
[It can be checked that $2, 3, 5, 7, 11, 101, 131, 151, 181, 191, 313, 353, 373, 383, 727, 757, 787, 797, 919, 929$, constitute the complete list of palindromic primes less than $2010$.]