The question asked here is a follow up to this question, which was answered by user GH from MO. [EDIT: The question is edited for clarification after receiving a comment on the original posting.]

As there, let $q$ be a natural number, let $P = \phi(q)/q$ be the "probability" that a randomly chosen integer is relatively prime to q. Then, following Montgomery and Vaughan in On the distribution of reduced residues, the second moment of the number of reduced residues modulo $q$ in an interval of length $h$ about its mean, $hP$, can be stated as $$ M_2(q;h) = qP^2 \sum_{\substack{{r \mid q }\\{r > 1}}} \mu(r)^2 \left( \prod_{\substack{ {p \mid q }\\{p \nmid r} }} \frac{p(p-2)}{(p-1)^2} \right) r^2 \phi(r)^{-2} \left\{ \frac{h}{r}\right\}\left( 1 - \left\{ \frac{h}{r}\right\}\right). $$ Applying the inequality $\{\alpha\}(1 - \{\alpha\}) \leq \alpha$ then gives the upper bound $$\tag{1} M_2(q;h)\leq qhP, $$ while a lower bound is given by $$ M_2(q; h) \geq qhP - qhPQ + O(qhP^2) $$ where $Q=\prod_{\substack{{p \mid q}\\{p>h}}} (1-1/p)$.

Since $\{\alpha\}\leq 1$, we also have that $\{\alpha\}(1 - \{\alpha\}) \leq 0.25$. This allows for a second upper bound, $$\tag{2} M_2(q;h) \leq 0.25 \, qP^2 \sum_{\substack{{r \mid q }\\{r > 1}}} \mu(r)^2 \left( \prod_{\substack{ {p \mid q }\\{p \nmid r} }} \frac{p(p-2)}{(p-1)^2} \right) r^2 \phi(r)^{-2}. $$

As pointed out by user Ofir Gorodetsky in the comment below, the upper bound in (1) is already optimal for those choices of $q$ such that the upper and lower bounds are of same order. For example, taking $q=\prod_{p\leq h^C} p$ for large enough $C$, we have that $M_2(q;h)\asymp qhP$.

Here I'd like to consider the rather opposite situation, when $C$ is small. Specifically, take $q=\prod_{p\leq h^C} p$, where $C<1$. Then we have that $Q=1$ and the lower bound takes the form $$ M_2(q; h) \geq O(qhP^2), $$ which leaves some more wiggle room between the upper and lower bounds, and I'd like to know whether the upper bound in (2) this case provides some improvement on (1).

# Question

Given $q=\prod_{p\leq h^C} p$, where $C<1$. Does the upper bound for $M_2(q;h)$ stated in (2) provide a sharper bound compared to (1) in this case? And is it possible to express the upper bound in (2) a bit more simplified or asymptotically?