The asymptotic frequency of square-free integers is known to be $6/\pi^2$, see [1].

Denote by $P_n$ the uniform distribution on $[1,n]$ and by $E_n$ the corresponding expectation. Then $$E_n(r)=\sum_{k \le \sqrt{n}} k P_n(r=k) \sim \sum_{k \le \sqrt{n}} k \cdot \frac{1}{k^2} \cdot\frac{6}{\pi^2} \sim \frac{3}{\pi^2} \log(n) \,,$$ where $A \sim B$ means that $A/B \to 1$ as $ n \to \infty$. Also, $$E_n(r^2)=\sum_{k \le \sqrt{n}} k^2 P_n(r=k) \sim \sum_{k \le \sqrt{n}} \frac{6}{\pi^2} \sim \frac{6n}{\pi^2} \,, $$ so the variance of $r$ is asymptotic to $6n/\pi^2$ as well.

[1] https://en.wikipedia.org/wiki/Square-free_integer

The asymptotic frequency of square-free integers is known to be $6/\pi^2$, see [1].

Denote by $P_n$ the uniform distribution on $[1,n]$ and by $E_n$ the corresponding expectation. Then $$E_n(r)=\sum_{k \le \sqrt{n}} k P_n(r=k) \sim \sum_{k \le \sqrt{n}} k \cdot \frac{1}{k^2} \cdot\frac{6}{\pi^2} \sim \frac{3}{\pi^2} \log(n) \,,$$ where $A \sim B$ means that $A/B \to 1$ as $ n \to \infty$. (In particular for $n=10^{10}$ the mean $E_n(r)$ is close to 7.) Also, $$E_n(r^2)=\sum_{k \le \sqrt{n}} k^2 P_n(r=k) \sim \sum_{k \le \sqrt{n}} \frac{6}{\pi^2} \sim \frac{6\sqrt{n}}{\pi^2} \,, $$ so the variance of $r$ is asymptotic to $6\sqrt{n}/\pi^2$ as well.

[1] https://en.wikipedia.org/wiki/Square-free_integer