Solution:
We claim that if the decimal expansion of $a$ has at least $2012$ digits, then $a$ contains the required substring. Let the decimal expansion of $a$ be $a_{k} a_{k-1} \ldots a_{0}$. For $i = 0, \ldots, 2011$, let $b_{i}$ be the number with decimal expansion $a_{i} a_{i-1} \ldots a_{0}$. Then by the pigeonhole principle, $b_{i} \equiv b_{j} \pmod{2011}$ for some $i < j \leq 2011$. It follows that $2011$ divides $b_{j} - b_{i} = c \cdot 10^{i}$. Here $c$ is the substring $a_{j} \ldots a_{i+1}$. Since $2011$ and $10$ are relatively prime, it follows that $2011$ divides $c$.