This problem can be solved using the binomial expansion. Firstly, we know $2011^{2011}\equiv 11^{2011}\pmod {1000}$. Then, rewrite $11$ as $(10+1)$ and apply the binomial expansion. $$(10+1)^{2011} = \cdots + C_{2011}^2\times 10^2 + C_{2011}^1 \times 10  + 1$$

It is obvious that all the previous terms where the power of $10$ is greater than $2$ always end with $000$, therefore can be safely discarded if we are just looking for the last three digits. Therefore,

$$C_{2011}^2\times 10^2 + C_{2011}^1 \times 10  + 1\equiv 11\times 10\times 50+110+1\equiv 611\pmod{1000}$$

Hence, the hundreds digit is $6$.