Answer: 52

Solution:

Measure time $t$ in seconds from when Al steps on. Al stands on the walkway, so he is at $6t;$ Bob moves at $6 + 4 = 10$ feet per second, so he is at $10(t - 2);$ Cy walks beside the walkway at $8$ feet per second, so he is at $8(t - 4).$ All three are moving once $t \ge 4.$

The middle person's position doubled must equal the sum of the other two. If Bob were in the middle, $20(t-2) = 6t + 8(t-4)$ gives $t = \frac{4}{3} \lt 4,$ impossible. If Cy were in the middle, $16(t-4) = 6t + 10(t-2)$ reduces to $-64 = -20,$ with no solution. If Al is in the middle, $12t = 10(t-2)$ $+ 8(t-4) = 18t - 52,$ so $t = \frac{26}{3}.$

At that moment Al is at $6 \cdot \frac{26}{3} = 52$ feet, while Bob and Cy are at $\frac{200}{3}$ and $\frac{112}{3},$ whose average is indeed $52.$ The middle person is $52$ feet from the start.