I didn't work it out fully, but I think I have a working method.
The area of a triangle is its semiperimeter * inradius. Therefore, any triangle with the same incircle has the same area. Let's adjust our triangle to an easier one to deal with, like an isosceles one. Let's adjust the triangle so that the 6 and 7 are equal while keeping the same incircle.
If we call the angle between the equal sides of our new triangle theta, the area of the triangle is (1/2)(sin theta)(x2). Here, x is the length of one of the equal sides. Similarly, using the original triangle, we find that the area is (1/2)(sin theta)(6)*(7). Thus, x = sqrt(42).
Let's find the area of the triangle in terms of the third side m (which isn't equal to either of the other sides) and its altitude n. By Pythagoras, m2 + n2 = 42. However, we also know that the area of the triangle is 4sqrt(5) by Heron's formula using the original side lengths, so m*n = 8sqrt(5). After some algebra (namely, adding two of the second equation to the first and taking the square root, the subtracting two of the second equation from the first and taking the square root gives us m+n=sqrt(42+16sqrt(5)) and m-n=sqrt(42-16sqrt(5)) from which we can easily find m and n), we conclude that the altitude n is sqrt(106)/2.
From here, it gets a lot easier. Imagine slicing off a new base between each of the circles in the diagram. The new triangles are clearly similar. We can find the new altitude to the furthest side of the second largest circle (h2) in terms of the altitude (h1, which we called n before) pretty easily.
*Note that the ratio of the inradius (sqrt(5)/2 because sr=area of triangle) to the height (sqrt(106)/2) will always be sqrt(5/106). Let's call the reciprocal of this value k.
(h2) = (h1) - 2(r1) = (h1) - 2(h1)(k) = (h1)(1-2k)
(h3) = (h1) - 2(r1) - 2(r2) = (h1) - 2(h1)(k) - 2(h2)(k) = (h1)(1-2k) - 2k((h1)(1-2k)) = (h1)*((1-2k)2)
Thus:
hN = (h1)((1-2k)N-1)
Multiplying each term by (1/k), squaring each term, multiplying by pi, and summing up, we get:
(pi/(k2))((h1)2 + (h2)2 + ... ) = (pi/(k2))((h1)2 + (h1)2*((1-2k)2) + ... ) = ((h12)*(pi)/(k2))(1/(1 - (1-2k)2))
We know the value of h1 and k, but it is ugly so I don't want to plug it in. I can't help but think there is a version of this problem with much nicer side lengths so that the answer is like 5 or something.