Denote by $\varphi$ Euler's totient function, and adopt I. M. Vinogradov's ($\ll$) asymptotic notation. Furthermore, let $$S(Q)=\sum_{Q<q\le 2Q}\varphi(q^2).$$
We have $$S(Q)\ll \sum_{Q<q\le 2Q}q^2\ll \sum_{Q<q\le 2Q} Q^2\ll Q^3.$$ This implies that $$\limsup_{Q\to\infty} \frac{S(Q)}{Q^3}$$ is finite.
To what number does the above $\limsup$ evaluate?
Similarly, we have $$S(Q)\gg \sum_{Q<q\le 2Q}\frac{q^2}{\log\log q}\gg \sum_{Q<q\le 2Q} \frac{Q^2}{\log\log Q}\gg \frac{Q^3}{\log\log Q}.$$ The linear regression I ran on $S(Q)$ v. $Q^3$ over $1\le Q\le 1000$ fits extremely well. This makes me think that $$\liminf_{Q\to\infty} \frac{S(Q)}{Q^3}$$ is also worthy of consideration.
To what does the above $\liminf$ evaluate?
Let $$p_k(x)=\sum_{n=1}^x\varphi(x^k)$$ for $k=1,2$.
It is known that
$$p_1(x) = \frac{3}{\pi^2}x^2+o(x^2).$$
Thus, $$p_2(x)=\sum_{n=1}^xn\varphi(n)=xp_1(x)-\sum_{n=1}^{x-1}p_1(n)=\frac{3}{\pi^2}x^3+o(x^3)-\frac{1}{\pi^2}(x-1)^3+o((x-1)^3)=\frac{2}{\pi^2}x^3+o(x^3).$$
Now,
$$S(x)=p_2(2x)-p_2(x)=\frac{14}{\pi^2}x^3 +o(x^3).$$
So,
$$\lim_{n\rightarrow\infty}\frac{S(x)}{x^3}=\frac{14}{\pi^2}$$