jmc

The hundreds digit of $2011^{2011}$ is the same as the hundreds digit of $11^{2011}.$

By the Binomial Theorem, $$\begin{array}{rl}11^{2011}& =(10+1{)}^{2011}\\ & =10^{2011}+\left(\genfrac{}{}{0ex}{}{2011}{1}\right)10^{2010}+\left(\genfrac{}{}{0ex}{}{2010}{2}\right)10^{2009}\\ & +\cdots +\left(\genfrac{}{}{0ex}{}{2011}{2008}\right)10^3+\left(\genfrac{}{}{0ex}{}{2011}{2009}\right)10^2+\left(\genfrac{}{}{0ex}{}{2011}{2010}\right)10+\left(\genfrac{}{}{0ex}{}{2011}{2011}\right).\end{array}$$Note that all the terms up to $\left(\genfrac{}{}{0ex}{}{2011}{2008}\right)10^3$ are divisible by 1000. Thus, the hundreds digit of the given number is the same as the hundreds digit of the number $$\begin{array}{rl}\left(\genfrac{}{}{0ex}{}{2011}{2009}\right)10^2+\left(\genfrac{}{}{0ex}{}{2011}{2010}\right)10+\left(\genfrac{}{}{0ex}{}{2011}{2011}\right)& =\frac{2011\cdot 2010}{2}\cdot 10^2+2011\cdot 10+1\\ & =202125611.\end{array}$$Hence, the hundreds digit is $\boxed{6}.$