It's so big

$\begin{aligned} S & = 100 + 100^2 + 100^3 + ...+ 100^{2011} \\ & = 10^2 + 10^4 + 10^6+...+2^{4018}+2^{4020}+10^{4022} \\ & = 10^{4022} + 10^{4020} + 2^{4018} + ... + 2^6 + 10^4 + 10^2 \\ & = 1\underbrace{00...000}_{\text{4022 0's}} + 1\underbrace{00...000}_{\text{4020 0's}} + 1\underbrace{00...000}_{\text{4018 0's}} + ... + 1\underbrace{000000}_{\text{6 0's}} + 1\underbrace{0000}_{\text{4 0's}} + 1\underbrace{00}_{\text{2 0's}} \\ & = \underbrace{101010...01010100}_{\boxed{4023} \text{ digits}} \end{aligned}$

Here number of digit only depend on last digit 100^(2011) ie 10^(4022) . Take log then number of digit will be log 10^(4022) +1 =4023