For answering this question we need to know mainly two theorems:
$\text{(1) Lagrange’s Theorem : If $H$ is a subgroup of some finite group $G$ then }$
$\text{$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$the order of $G$ is divisible by the order of $H$.}$
$\text{(2) If a group has a prime order then it is cyclic.}$
Now it is given that $G$ is a group of order $6$. Since $6$ is not a prime number, so $G$ may or may not be cyclic [Since Theorem $(2)$ says being prime is just a sufficient condition for a group to be cyclic but not necessary]
Also, it is that that order of $H$ can be $2,3,4,5$ [as per the question $1<|H|<6$]. Now Theorem $1$ says that it is necessary for the order of a group to be divisible by the order of the subgroup.
Hence the only feasible orders of $H$ are $2,3$. [The orders $4,5$ are ruled out since they do not divide $6$.]
Now what we see is that the possible orders of $H$ are prime and hence from Theorem $2$, $H$ has to be cyclic.
So option $(B)$ is the correct option.