If $2^n=(... \underbrace{aa...aaa}_{x\text{ times}})_{10}$ for positive integer $n$ then we have
$$2^n=(... \underbrace{aa...aaa}_{x\text{ times}})_{10}=(b\underbrace{aa...aaa}_{x\text{ times}})_{10}=10^x b+a(10^{x-1}+...+1)$$ Where $a$ is a digit between $0$ and $9$ and $b$ is a natural number.
Simplifying more gives $$9 \times2^n=9b \times 10^x + a (10^x-1)$$
Now if $n>3$ and $x>3$ then taking mod 2 gives $v_2(a)>3$ which is impossible. So one $x$ or $n$ is not greater than $3$. But the book says answer is $x=125$. Am I interpret this wrong?
If $n\ge 4$, then $2^n$ must be a multiple of $16$, therefore so are its last four digits. There are no factors of $5$ so the last four digits cannot be $0000$. Now, which other possibilities for four identical last digits will give a multiple of $16$? Try them out and see.
The comments identify $2^{39}=...888$. If you work out the part above you are left with what must be the answer.