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 $ \mathop{\text{lcm}}[8m,10^{10}] = \mathop{\text{lcm}}[2^3m,2^{10}5^{10}] $, so the exponent of $ 2 $ in the prime factorization of $ \mathop{\text{lcm}}[8m,10^{10}] $ is equal to $ \max\{3+\alpha,10\} $. Similarly, the exponent of $ 2 $ in the prime factorization of $ 4\cdot\mathop{\text{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} $.