$b_m=v_2(m!)$. By well-known relation: $v_2(m!)=m-s_2(m)$, where $s_2(m)$ is sum of digits in binary base. Hence, $1990=m-b_m=s_2(m)$. Because in binary base only digits are $1$'s and $0$'s, it's obvious that smallest such $m$ is $2^1990-1$ Q.E.D.

Shouldn't it be 2^1990 + 2^1989 + ..... + 2^2 + 2 ? Clearly the smallest number ( in binary ) containing 1990 1's is 11111...1111 ( 1990 times)..