For any non-zero natural integer $n$, we set
$$\omega(n) = \operatorname{Card}\{p \text{ prime} : p \mid n\} = \sum_{\substack{p \mid n \\ p \text{ prime}}} 1.$$
Conclude that $\frac{1}{x}\left(\sum_{n \leqslant x} \left(\omega(n) - \ln_2(x)\right)^2\right) \underset{x \rightarrow +\infty}{=} O\left(\ln_2(x)\right)$.