Answer: C

Solution:

Let the order of the numbers be $$a, b, c, d, e.$$

Then the iterative average is $$\dfrac{\dfrac{\dfrac{\dfrac{a + b}{2} + c}{2} + d}{2} + e}{2}$$$$= \dfrac{a + b + 2c + 4d + 8e}{16}.$$

To minimize this, we make the order $$5, 4, 3, 2, 1,$$ which gives us a sum of $$\dfrac{5 + 4 + 6 + 8 + 8}{16} = \dfrac{31}{16}.$$

To maximize it, we have to reverse this order to get an average of $$\dfrac{1 + 2 + 6 + 16 + 40}{16} = \dfrac{65}{16}.$$

The difference between these is $$\dfrac{65}{16} - \dfrac{31}{16} = \dfrac{34}{16} = \dfrac{17}{8}.$$

Thus, C is the correct answer.