I tried this way, I only need to know if this is correct or if there are better ways to solve this:
$2^{1000}$ does not have a factor of $5$ obviously therefore we can assume
$ 10^{m} < 2^{1000} < 10^{m+1}$ for some $m$
Assume $ k = 2^{1000}$, then take log on both sides $\log k = 1000 \log 2 \approx 301.02999 > 301$
Therefore $2^{1000}$ has $302$ digits.