Let $C$ be the total capacity of the barrel, in liters.
1. Set up the relationship

The problem states that adding $k$ liters changes the fraction of the barrel that is full from $\(\frac{1}{5}\)$ to $\(\frac{2}{3}\)$. The difference between the final volume and the initial volume is $k$ liters.

$$
\(\text { Final Volume }- \text { Initial Volume }=k\)
$$


$$
\(\left(\frac{2}{3} \times C\right)-\left(\frac{1}{5} \times C\right)=k\)
$$

2. Combine the fractions

To subtract the fractions, we need a common denominator, which is $\(3 \times 5=15\)$ :

$$
\(\begin{aligned}
& \frac{2}{3}=\frac{2 \times 5}{3 \times 5}=\frac{10}{15} \\
& \frac{1}{5}=\frac{1 \times 3}{5 \times 3}=\frac{3}{15}
\end{aligned}\)
$$


Substitute the equivalent fractions back into the equation:

$$
\(\begin{gathered}
\frac{10}{15} C-\frac{3}{15} C=k \\
\left(\frac{10}{15}-\frac{3}{15}\right) C=k \\
\frac{7}{15} C=k
\end{gathered}\)
$$

3. Solve for the capacity ( $C$ ) in terms of $k$

To find $C$, multiply both sides of the equation by the reciprocal of the fraction $\(\frac{7}{15}\)$, which is $\(\frac{15}{7}\)$ :

$$
\(\begin{gathered}
C=k \times \frac{15}{7} \\
C=\frac{15 k}{7}
\end{gathered}\)
$$


The capacity of the barrel is $\(\frac{15}{7} k\)$ liters.

This matches the option \(15/7k\).