To get you started some rather basic (or even trivial, depends) observations, putting $\,n_p=$ number of Sylow $p-$subgroups , $\,G:=\,$ simple group of order $\,168\,$:
$$168=2^3\cdot 3\cdot7\Longrightarrow n_7=8\;,\;n_3=7,28$$
It can't be $\,n_3=4\,$ as then $\,G\,$ has a subgroup of index 4, which is impossible since this would mean $\,G\,$ is isomorphic with a subgroup of $\,S_4\,$ . For the same reason, it must be $\,n_2=7,\,21\,$...
Perhaps reading here you'll have have the whole view. Don't worry about the number of pages as the first ones are basic results listed.
As commented back in the day, the OP's solution is correct. Promoting a modified (IMHO simplified) form of the idea from Thomas Browning's comment to an alternative answer.
The group $G$ contains $23\cdot24-24\cdot22=24$ elements outside the Sylow $23$-subgroups. Call the set of those elements $X$. Clearly $X$ consists of full conjugacy classes. If $s\in G$ is an element of order $23$, then consider the conjugation action of $P=\langle s\rangle$ on $X$. The identity element is obviously in an orbit by itself. By Orbit-Stabilizer either $X\setminus\{1_G\}$ is a single orbit, or the action of $P$ has another non-identity fixed point in $X$.
- In the former case all the elements of $X\setminus\{1_G\}$ must share
the same order, being conjugates, in violation of Cauchy's theorem
implying that there must exist elements of orders $2$ and $3$ in $X$
at least.
- In the latter case $P$ is centralized, hence normalized, by a non-identity element $\in X$. But this contradicts the fact that $P$ is known to have $24=[G:P]$ conjugate subgroups, and hence must be self-normalizing, $N_G(P)=P$.
Best Answer
If $n_3=4$ you have $8$ elements of order $3$. so there are $4$ elements of other orders. Those are exactly enough to have a subgroup of order $4$, so it is unique. Hence normal.