jmc

Let $\alpha$ be the exponent of $2$ in the prime factorization of $m$. That is, $m=2^{\alpha }\cdot t$, where $t$ is some odd integer.

Note that $\mathrm{lcm}[8m,10^{10}]=\mathrm{lcm}[2^{3}m,2^{10}{5}^{10}]$, so the exponent of $2$ in the prime factorization of $\mathrm{lcm}[8m,10^{10}]$ is equal to $\max \{3+\alpha ,10\}$.

Similarly, the exponent of $2$ in the prime factorization of $4\cdot \mathrm{lcm}[m,10^{10}]$ is $2+\max \{\alpha ,10\}$. Thus we have $$\max \{3+\alpha ,10\}=2+\max \{\alpha ,10\},$$which is possible only if $\alpha =9$. So, $m$ is divisible by $2^9=512$. The only 3-digit multiple of $2^9$ is $512$ itself, so $m=\boxed{512}$.