Baltic Way 2019

Yes. Let us observe powers of two of the form $2^{10k}=(2^{10}{)}^k=1024^k=(1,024{)}^k\cdot 10^{3k}$. Let $m$ be the smallest integer such that $(1,024{)}^m\ge 3,9$ ($m$ must exist since $(1,024{)}^k$ is an exponential function and is unbounded from above). From the definition of $m$ we know that $(1,024{)}^{m-1}<3,9$, therefore $(1,024{)}^m=(1,024{)}^{m-1}\cdot 1,024<3,9\cdot 1,024=3,9+3,9\cdot 0,024<3,9+4\cdot 0,025=4,0$. Thus $3,9\le (1,024{)}^m<4,0$ and the first digits of $2^{10m}=(1,024{)}^m\cdot 10^{3m}$ are indeed 3 and 9.

Remark 1. The value of $m$ found above is 58 and the respective power of two is $2^{580}=3,\mathrm{957286...}\cdot 10^{174}$.

Remark 2. The smallest power of two satisfying the requirements is $2^{95}=39614081257132168796771975168$.