III OBM

Take $n$ to be the smallest integer such that $2^n\ge 10^{m-1}$. Then $2^{n-1}<10^{m-1}$, so $2^{n+2}<8\cdot 10^{m-1}<10^m$. So $2^n$, $2^{n+1}$ and $2^{n+2}$ all have $m$ digits. Thus there are at least $3$ powers of $2$ with $m$ digits.

$2^{n-1}\ge 5\cdot 10^{m-2}$ (otherwise $2^n<10^{m-1}$). Hence $2^{n+4}=32\cdot 5\cdot 10^{m-2}>10^m$, so $2^{n+4}$ has more than $m$ digits. Thus there are at most $4$ powers of $2$ with $m$ digits.

There are $4$ if there is an integer between $\frac{m-1}{{\log }_{10}2}$ and $\frac{m}{{\log }_{10}2}-3$.