By using some more powerful results, it is possible to do this a lot easier.
The two main ingredients for this will be the following:
Burnside's $pq$-Theorem: If only two distinct primes divide the order of $G$, then $G$ is solvable.
Burnside's Transfer Theorem: If $P$ is a $p$-Sylow subgroup of $G$ and $P\leq Z(N_G(P))$ then $G$ has a normal $p$-complement.
So when looking for a non-abelian simple group, the first result immediately tells us that we can assume at least $3$ distinct primes divide the order of $G$.
The way we will use the second result is the following:
Let $P$ be a $p$-Sylow subgroup where $p$ is the smallest prime divisor of $|G|$. We will show that if $|P| = p$ then $P\leq Z(N_G(P))$ (and then the above result says that $G$ is not simple).
To see this, we note that $N_G(P)/C_G(P)$ is isomorphic to a subgroup of $\rm{Aut}(P)$ (this is known as the N/C-Theorem and is a nice exercise), which has order $p-1$, and since the order of $N_G(P)/C_G(P)$ divides $|G|$, this means that $N_G(P) = C_G(P)$ and hence the claim (since we had picked $p$ to be the smallest prime divisor).
The same argument actually shows that if $p$ is the smallest prime divisor of $|G|$, then either $p^3$ divides $|G|$ or $p = 2$ and $12$ divides $|G|$. The reason for this is that if the $p$-Sylow is cyclic of order $p^2$ the precise same argument carries through, since in that case all prime divisors of $|\rm{Aut}(P)|$ are $p$ or smaller.
If $P$ is not cyclic then the order of $\rm{Aut}(P)$ will be $(p^2-1)(p^2 - p) = (p+1)p(p-1)^2$ and the only case where this can have a prime divisor greater than $p$ is when $p = 2$ in which case that prime divisor is $3$, and we get the claim.
In summary we get that the order of $G$ must be divisible by at least $3$ distinct primes, and either the smallest divides the order $3$ times, or the smallest prime divisor is $2$ (actually, by Feit-Thompson, we know this must be the case, but I preferred not to also invoke that), and $3$ must also divide the order.
This immediately gives us $60$ as a lower bound on the order of $G$, and the next possible order would be $2^2\cdot 3\cdot 7 = 84$ and all further orders are greater than $100$. So we are left with ruling out the order $84$ (which you have done), and showing that the only one of order $60$ is $A_5$.
(It's not true that any cyclic subgroup of a group is normal. You can see this since for example the symmetric group $S_3$ has a cyclic subgroup $\{e,(12)\}$ of order 2. It is not normal because $(23)(12)(23)^{-1}=(13)$)
By Lagrange's theorem, the non-trivial proper subgroups have order 3 or 7.
As you have correctly identified, from Sylow's theorems, $G$ has a unique subgroup $N$ of order 7. It must be normal, since for any prime number $p$ the Sylow $p$-subgroups of a group form a single conjugacy class of subgroups.
Suppose (for contradiction) that it also has a normal subgroup $K$ of order 3. Then $N \cap K =\{e\}$ (By Lagrange's theorem, the order of $N \cap K$ divides 3 and 7, so is 1). Their product $NK$ is thus the whole group $G$, since it has order $\frac{|N||K|}{|N \cap K|}=\frac{3\times7}{1}=|G|$. (To see this consider the map $f:N\times K \to G, (n,k)\mapsto nk$.) So any $n \in N$ and $k \in K$ commute. (Consider an element of the form $nkn^{-1}k^{-1}$. It is in $N\cap K$, so is $e$.)
We would thus have $G$ is isomorphic to the direct product of $N$ and $K$. In particular $G$ would be abelian, which is a contradiction.
So $G$ has no normal subgroup of order 3, and by Sylow's theorems has only one of order 7. Hence option 2 is correct.
Best Answer
If $n_3=13$, then there should be $13*(3-1)=26$ elements of order $3$. This is because there are $13$ distinct subgroups of order $3$ while each subgroup has $2$ elements of order $3$. Also, note that each pair of subgroups of order $3$ intersect trivially.
If $n_3=1$, then $G$ has a normal subgroup of order $3$,say $P$.
Note that $G$ has also a normal subgroup of order $13$,say $Q$.
Hence $G=P\times Q$ which is abelian since $P$ and $Q$ are both abelian, a contradiction.
So $n_3$ must be $13$ and the number of elements of order $3$ are $26$.