combinations

We assume all number less than $1000$ are $3$ digit number. $1$ -digit and $2$ -digit number can be represented as $3$ -digit number by adding $0$ (or 0's) at first. Now, there are total $1000\times 3=3000$ digits. By symmetry all digits( $0$ to $9$ ) are equal in number. So there are $\frac{3000}{10}=\fbox{300}$

In general, if you write all numbers from 1 to $10^{ n }$ , 5 will be written $n({ 10 }^{ n-1 })$ times.