Find the $26^{th}$ digit of a $50$ digit number divisible by $13$.

$10^{50}$ is a 51-digit number. And in a 50-digit number, the digit 26th from the left is represented by $10^{24}$.

Other than these two mistakes, I find your approach entirely reasonable. And if they were looking for a 51-digit number, with all except the 25th digit from the left being $1$, then it would've been correct too.

Edit: After having corrected these two off-by-one errors, the solution looks fine.

Another way is to use the trick from Wikipedia (that doesn't solve your solution)

Taking $N$ from the right, and applying the sequence $(1, −3, −4, −1, 3, 4)$ as instructed on the page (multiply the digits from the right by the given numbers in sequence), we get

$0$ for the 6 first digits from the right ($1-3-4-1+3+4=0$), repeating the sequence, $0$ up to digit 24 (from right), we still have $0$

Then, the next $6$ are our $a$ and $5\times 1$, or $$a-3-4-1+3+4\\=a-1$$

We did $30$ digits, $20$ to go. The next $18$ will give $0$, the last $2$ give $1-3$, thus the whole sum is $$a-1-2=a-3$$ The only digit that would have $a-3\equiv 0\pmod {13}$ is $$\bbox[5px,border:2px solid #ba9]{a=3}$$