The given number has $11$ digits. There are $2^{11} - 1 = 2047$ positive integers with at most $11$ digits each of which is either $0$ or $1$. We solve the problem by subtracting the number of positive integers that are not less than the given number.
The $11$-digit numbers larger than the given number are all of the form $11111abcde$. There are $2^5 = 32$ numbers in this form. Among them, $4$ numbers $1111100000$, $1111100001$, $1111100010$ and $1111100011$ are less than the given number. Thus the number of positive integers consisting of zeros and ones and being less than $1111100100$ is $2047 - 32 + 4 = 2019$.

Ordering any two digit sequences that constitute a positional representation on different bases does not depend on the base. This means that if a number is larger than another number on base $2$ then the first number is larger than the second number also on base $10$. The number $1111100100$ on base $2$ equals $2^{10} + 2^9 + 2^8 + 2^7 + 2^6 + 2^5 + 2^2 = 2020$ on base $10$. Hence, to solve the problem, it suffices to count all positive integers less than $2020$. The result is obviously $2019$.