Let $M$ be a subgroup of $\mathbb Z^3$ such that $\mathbb Z^3/M$ is cyclic of order $k$. Then $M$ is free of rank $3$, so there is a matrix $A\in M_3(\mathbb Z)$ of non-zero determinant such that $M=A\cdot\mathbb Z^3$. Then there exists $3\times 3$ matrices $P$ and $Q$, invertible over $\mathbb Z$, such that $PAQ$ is a diagonal matrix with entries $a$, $b$, $c$ such that $a\mid b\mid c$, and the three numbers $a$, $b$, $c$ depend only on $M$.

It follows that there is an element $g\in\mathrm{SL}(3,\mathbb Z)$ such that $g(M)$ is generated by $(a,0,0)$, $(0,b,0)$ and $(0,0,c)$. Now, since $\mathbb Z^3/g(M)$ is cyclic of order $k$, we must have $a=b=1$, $c=k$.