Given that,
DIAGRAM
In drum$1,$ $A$ and $B$ are in the ratio of $18 : 7.$
So, the quantity of $A = \frac{18}{25}$
The quantity of $B = \frac{7}{25}$
In final mixture, $A$ and $B$ are in the ratio of $13 : 7.$
So, the quantity of $A = \frac{13}{20}$
The quantity of $B = \frac{7}{20}$
In drum$2,$ let us assume the proportion of $B$ with respect to the overall mixture is $x.$
So, $ \frac{ \frac{7}{25} \times 3 + x \times 4}{7} = \frac{7}{20}$
$\Rightarrow \frac{21}{25} + 4x = \frac{49}{20}$
$\Rightarrow 20 (21 + 100x) = 25 \times 49$
$\Rightarrow 420 + 2000x = 1225$
$\Rightarrow 2000x = 805 $
$ x = \frac{805}{2000} $
$ \boxed{x = \frac{161}{400}} $
In drum$2, \; B$ is $\frac{161}{400} $
So, $A$ should be the remaining.
Thus, $A$ is $\frac{239}{400}. $
$\therefore$ In drum$2, \; A$ and $B$ are in the ratio of $239 : 161.$
Correct Answer$: \text{C}$