$2007_{10} = 2202100_{3}$
All numbers of six or less digits in base 3 have been written.
The form of each palindrome is as follows
1 digit - $a$
2 digits - $aa$
3 digits - $aba$
4 digits - $abba$
5 digits - $abcba$
6 digits - $abccba$
Where $a,b,c$ are base 3 digits
Since $a \neq 0$, this gives a total of $2 + 2 + 2\cdot 3 + 2\cdot 3 + 2\cdot 3^2 + 2\cdot 3^2 = 52$ palindromes so far.
7 digits - $abcdcba$, but not all of the numbers are less than $2202100_{3}$
Case: $a=1$
All of these numbers are less than $2202100_3$ giving $3^3$ more palindromes
Case: $a=2$, $b\neq 2$
All of these numbers are also small enough, giving $2\cdot 3^2$ more palindromes
Case: $a=2$, $b=2$
It follows that $c=0$, since any other $c$ would make the value too large. This leaves the number as $220d022_3$. Checking each value of d, all of the three are small enough, so that gives $3$ more palindromes.
Summing our cases there are
$52 + 3^3 + 2\cdot 3^2 + 3 = 100 \Rightarrow \mathrm{(A)}$