6. (10 points) There are two docks, $A$ and $B$, on a river, with $A$ upstream and $B$ downstream. Two people, A and B, start from $A$ and $B$ respectively at the same time, rowing towards each other, and meet after 4 hours. If A and B start from $A$ and $B$ respectively at the same time, rowing in the same direction, B catches up with A after 16 hours. It is known that A's rowing speed in still water is 6 kilometers per hour, then B's rowing speed in still water is $\qquad$ kilometers per hour.
【Analysis】In the problem of boats traveling in a stream, the speed at which two boats meet is the sum of their speeds, and the speed at which one boat catches up with the other is the difference in their speeds.

When two boats travel towards each other, the distance they cover is the distance between docks A and B. When they travel in the same direction, the distance difference between the two boats is also the distance between docks A and B. Therefore, based on the equal distance, let the speed of boat B be \( x \) kilometers per hour, and set up the equation \((x+6) \times 4 = (x-6) \times 16\) to solve the problem.

【Solution】Solution: Let the speed of boat B be \( x \) kilometers per hour,
\[
\begin{aligned}
(x+6) \times 4 & = (x-6) \times 16 \\
4x + 24 & = 16x - 96 \\
12x & = 120 \\
x & = 10
\end{aligned}
\]

Answer: Boat B travels 10 kilometers per hour in still water.
Therefore, the answer is: 10.