Easy Peasy

First a fall , we will take an example. Number of digits in $\large 100+100^2 = 10100\implies\boxed{5}$ Which is equal to the number of digits in $\large100^2 =10000\implies\boxed{5}$

---

So the number of digits in $\large100+100^2+100^3=1000000+10000+100=1010100\implies\boxed{7}$ Which is again equal to the number of digits in $\large100^3=1000000\implies\boxed{7}$

---

Therfore we get that the number of digits in $(100^x+100^{x-1}+100^{x-2}+.....100^2+100^1)$ is equal to the number of digits in $(100^x)$

$\therefore$ Number of digits in $(100^{2011}+100^{2010}+.....+100^2+100)$ = Number of digits in $100^{2011}$ = $\large\color{#3D99F6}{\text{4023}}$

---