Suppose that a cyclic group $G$ has exactly three subgroups: $G$ itself, $\{ e \}$, and a subgroup of order $7$. What is $|G|$? What can you say if $7$ is replaced with $p$ where $p$ is prime?
I know that $7$ is a divisor of $|G|$ here, but I can't figure anything else past that.
On
A finite cyclic group has a subgroup of order $d$ for every divisor $d$ of $|G|$. So we are told that $|G|$ has exactly three divisors, $1$, $p$, and $|G|$. This implies $|G|=p^2$. (In fact, we immediately see that $|G|$ must be a perfect square because all non-squares have an even nmumber of divisors).
Recall that subgroups of a finite cyclic group $G$ are in one-to-one correspondence with divisors of $|G|$.
So, you need a number $n$, with its only divisors being $1$,$7$ and $n$. Obviously $n=49$, and $G=C_{49}$ - the cyclic group of order $49$.