A careful analysis of the answer to the linked question will give an asymptotic result instead of an upper bound.

Write $\left( \frac{n}{\phi(n)} \right)^a$ as $\prod_{p \mid n}(1+g(p))=\sum_{d \mid n} g(d)\mu^2(d)$, where $g$ is completely multiplicative and given by $g(p)=\left(1 + \frac{1}{p-1} \right)^a - 1$. Note that
$$g(p) \le \frac{2^a-1}{p-1},\quad g(d) \le \frac{\mu^2(d)}{d} (2^{a+1}-2)^{\Omega(d)}.$$

The sum becomes
$$\sum_{d \le x} \lfloor \frac{x}{d} \rfloor g(d)\mu^2(d) =x\sum_{d \le x} \frac{g(d)\mu^2(d)}{d} + O\left( \sum_{d\le x} g(d)\mu^2(d) \right) $$
$$\qquad =x\sum_{d} \frac{g(d)\mu^2(d)}{d} + O\left( \sum_{d\le x} g(d)\mu^2(d) \right) + O\left( x\sum_{d>x} \frac{g(d)\mu^2(d)}{d} \right).$$

- Finish by noting that $\sum_{d\le x} g(d)\mu^2(d)$ is $o(x)$, and that $\sum_{d>x } \frac{g(d)\mu^2(d)}{d}$ is $o(1)$ (so in particular $C(a):=\sum_{d } \frac{g(d)\mu^2(d)}{d}=\prod_{p} \left( 1+\frac{g(p)}{p}\right)$ converges):
$$\Omega(d) = O(\frac{\ln d}{\ln \ln d}) \implies$$
$$\sum_{d\le x} g(d)\mu^2(d) \le \sum_{d\le x} \frac{(2^{a+1}-2)^{O(\frac{\ln d}{\ln \ln d})}}{d} =\sum_{d\le x} O_a\left( \frac{1}{\sqrt{d}} \right) = O_a(\sqrt{x}),$$
$$\sum_{d>x} \frac{g(d)\mu^2(d)}{d} \le \sum_{d>x} \frac{(2^{a+1}-2)^{O(\frac{\ln d}{\ln \ln d})}}{d^2} =\sum_{d>x} O_a\left( \frac{1}{d\sqrt{d}} \right) = O_a(\frac{1}{\sqrt{x}}).$$

See Chapter 2.4 in these notes for similar examples.

Remark: The estimate $\Omega(d) = O\left( \frac{\ln d}{\ln \ln d} \right)$ is elementary - it relies only on the Chebyshev estimates. In fact, one only needs $\Omega(d) = o(\ln d)$, which follows from the fact that $\pi(d) = o(d)$ (zero-density of primes), a classical result already due to Legendre.