Show that the permutation $(1 \space 2 \space 3)$ can not be a cube of any element of $S_n.$

Hint, as requested: the order of an element $\sigma$ in $\mathfrak{S}_n$ is the least common multiple of the cycle lengths of $\sigma$.

Your mistake happened when you declared $o(a)=9$. That's not true.

It should read "$|a|$ divides 9". (If you don't mind, I would like to use $|a|$ notion instead of $o(a)$ for order of $a\in S_n$.)

This means $|a|\in \{1,3,9\}$.

Let $|a|=1$. That means $a=\varepsilon$ and hence $(123)=a^3=\varepsilon^3=\varepsilon$ which is contradiction.

Let $|a|=3$. Then $a^3=\varepsilon=(123)$, contradiction again.

And if $|a|=9$ (that means $n\geqslant 9$ otherwise it will be impossible) we can take $a=(a_1a_2\cdots a_9)$ so that $a^3=(a_1a_4a_7)(a_2a_5a_8)(a_3a_6a_9)$. Contradiction once again.