Solution:

Let $k$ be in $X$. There are $7!\,6!\,(8-k)$ choices of the sequences $a_{1}, a_{2}, \ldots, a_{7}$ and $b_{1}, b_{2}, \ldots, b_{7}$ for which $2k-1$ is an $a_{j}$ with $a_{j}<b_{j}$. Indeed, the $a$'s can be any of the $7!$ permutations of the 7 odd integers; then $j$ is the subscript such that $a_{j}=2k-1$, and $b_{j}$ must be one of the $8-k$ numbers in $\{2k, 2k+2, \ldots, 14\}$; the remaining 6 even integers can be arranged in $6!$ ways.

The total of the payments by $B$ to $A$ for the $(7!)^{2}$ possible rounds is then
$$\sum_{k=1}^{7} 7!\,6!\,(8-k) = 7!\,6!\,(7+6+\cdots+1) = 7!\,6!\,28 = (7!)^{2} 4$$
$A$ pays to $B$ a total of $(7!)^{2}c$; so $c=4$ makes the game fair.

Solution:

For those who know that the sum of expected values (averages) is the expected value of the sum, we note that in the first spot, $1, 3, 5, 7, 9, 11, 13$ are equally likely, and hence the average payment is the average of the payments for each of these numbers, $7/7, 6/7, 5/7, 4/7, 3/7, 2/7, 1/7$. Hence in the first spot the average payment is
$$\frac{(7/7 + 6/7 + 5/7 + 4/7 + 3/7 + 2/7 + 1/7)}{7} = 4/7 \text{ of a cent. }$$
Since there are 7 spots, with an average payment of $4/7$ cent each, the total payment averages 4 cents; so $c=4$ makes the game fair.