Let $'a'$ be an element in $G$ of order $5$.
Then the subgroup $\langle{g\rangle}$ generated by $'a'$ is a cyclic group of order $5.$
That is, $\langle{a\rangle} = \{e,a,a^{2},a^{3},a^{4}\}$, where $'e'$ is the identity element in $G.$
Note that the order of each non-identity element in $\langle{a\rangle}$ is $5.$
Also, if $'b'$ is another element in $'G'$ of order $5$, then we have either $\langle {a} \rangle = \langle {b\rangle}$ (or) $\langle {a} \rangle \cap \langle {b} \rangle=\{e\}$.
This follows from the fact that the intersection $\langle {a\rangle} \cap \langle {b\rangle} $ is a subgroup of the order $5$ group $\langle {a \rangle},$ and thus
the order of $\langle {a \rangle } \cap \langle {b\rangle} $ is either $5$ (or) $1$.
On the other hand, if $H$ is a subgroup of $G$ of order $5$, then every non-identity element in $H$ has order $5$.
These observations imply that each subgroup of order $5$ contains exactly $4$ elements of order $5$ and each element of order $5$ appears in exactly one of such subgroups.
As there are $48$ elements of order $5$, there are $\dfrac{48}{4}=12$ subgroups of order $5$.
So, the correct answer is $(A)$.
