For any such $X$ we have $X\le_mTh(\mathbb{N})$ but not conversely (and in fact $Th(\mathbb{N})$ is vastly more complicated than $X$).
To see that $X\le_mTh(\mathbb{N})$, we want to translate a question of the form "Is $n\in X$?" into one of the form "Is $\psi_n$ a true sentence?" for some appropriate sentence $\psi_n$. Now "$n\in X$" is equivalent to "$\gamma(n)$," and that's almost a sentence in the language of arithmetic - the only issue is that "$n$" isn't a symbol in the language of arithmetic (which consists only of $0,1,+,\times,<$). Do you see a quick way around this?
Each natural number $n$ has a corresponding numeral, usually denoted "$\underline{n}$," which is just an appropriate string of the symbols $0$," "$1$," "$($," and "$)$" (or is the symbol "$0$" if $n$ is $0$): e.g. $\underline{4}$ is the string $((1+1)+1)+1$. Then if we let $\psi_n=\gamma(\underline{n})$, we have that $\psi_n\in Th(\mathbb{N})$ iff $n\in X$.
Note that here I'm conflating e.g. the number $0$ and the constant symbol $0$; this is annoying, but standard.
The other direction is a consequence of Tarski's undefinability theorem: just show that anything many-one reducible (or even Turing-reducible) to an arithmetically definable set is itself arithmetically definable. This will show that we never have $Th(\mathbb{N})\le_mX$ (or even $\le_TX$) for such an $X$. The key here is to show that the basic notions of computability theory are arithmetically definable:
Use Kleene's $T$-predicate to talk about computable reductions.
[Newbie not sure about] the best way to get help
There is a help centre including How do I ask a Good Question? My favourite advice is Picture yourself addressing a peer busy with something else. $\endgroup$