Is the group $|G|=3^3\cdot 5\cdot 7$ possible?
I've been examining it, and it seems like there can't be enough elements in the Sylow-$p$-subgroups.
That is, there can only be $21$ (or $1$) groups of order $5$ (the only number that is equal to $1 + 5k$ and divides $189$). We also know they can't contain a proper subgroup, so their intersection must be trivial. This implies there are $84$ elements of order $5$, at most.
Similarly, the number of $7$-sylow subgroups can only be $1$ or $15$, and again the intersection is trivial. So, at most $90$ elements are of order $7$.
This leaves elements of order $3$, $9$, or $27$. The number of Sylow subgroups is $1$ or $7$, so the most amount elements can be found if there are seven Sylow $3$ groups which intersect non-trivially, which means there are at most $182$ elements of this type.
But $1+182+84+90 \neq 945$.
Consider the cyclic group of order $3^3\cdot 5\cdot 7$.
In fact, for any $n\in\Bbb N$, there exists at least one group, namely the cyclic group of order $n$.