The easiest method is to simply count palindromes of length exactly n. For even n, this is $26^{n/2}$, and for odd n, $26^{(n-1)/2}$.
So, if we include lengths 1 and 2, we want to add $26 +26 +26^2 +26^2 + 26^3 + 26^3 + \ldots$
If n is even, this is just twice the sum of a geometric series. If n is odd, you get one extra term to add at the end. (You can subtract the 52 sequences of length <= 2 at the end if you like).
If the leading bit can't be zero, then neither can the last bit. So you can represent an $11$-bit binary palindrome as follows:
$$1\ b_1\ b_2\ b_3\ b_4\ x\ b_4\ b_3\ b_2\ b_1\ 1$$
where $x$ can be either $0$ or $1$, and $b_1b_2b_3b_4$ is any $4$-bit number. It follows that there must be $32$ numbers of this form. A $10$-bit palindrome will follow the same pattern, but without the $x$ in the middle, so there are sixteen $10$-bit palindromes. Similarly, for $9$-bit and $8$-bit palindromes, there are $16$ and $8$ possibilities respectively.
So the total number of palindromes in the range $2-2027$ (i.e., with $2–11$ bits inclusive) must be $32+16+16+8+8+4+4+2+2+1$, which is $93$.
An even easier method:
It just occurred to be that there's another way of looking at this problem. If you ignore the requirement that the leading digit must be $1$, then the entire set of palindromes reduces to:
$$b_1\ b_2\ b_3\ b_4\ b_5\ x^\prime\ b_5\ b_4\ b_3\ b_2\ b_1$$
where $x^\prime$ can be $1$, $0$ or nothing. If there are any leading zeroes, these are matched by an equal number of trailing zeroes. If we suppose that all the palindromes are post-processed by stripping the leading and trailing zeroes (e.g., by converting $00011011000$ to $11011$), then we can obtain the same result by multiplying the number of permutations of $b_1b_2b_3b_4b_5$ ($31$, because we are still excluding $00000$) by the number of possible values of $x^\prime$ ($3$)
Best Answer
Suppose $m$ is a palindrome. Any positive binary integer has first digit $1$, so $m$ has last digit $1$, and if $m$ has $k$ digits and $k > 1$, this leaves $k - 2$ digits. If $k$ is even, so is $k - 2$, and since $m$ is a palindrome, the (free) choice of the first $\frac{1}{2}(k - 2)$ of the remaining digits determines the last $\frac{1}{2} (k - 2)$ of the remaining digits, and so there are $2^{(k - 2) / 2}$ binary palindromes of even length $k$.
Similarly, if $k$ is odd, the free choice of the first $\frac{1}{2}(k - 1) $ digits determines the last $\frac{1}{2} (k - 3)$ of the digits. and so there are $2^{(k - 1) / 2}$ binary palindromes of odd length $k$.
Of course, in the special case that $k = 1$ there is just one palindrome, namely $1$ itself (and so the above general formula for odd $k$ extends to this case too).