d2-d1 is TWICE the height of the string above the land. If you look at it from the raidus’ point of view (which is half the diameter):

r2-r1 is the real height of the string above the land:

r2-r1=h

if 2*r2=d2 and 2r1=d1, by substitution we can get:

d2/2 – d1/2 = h

to simplify:

(d2-d1)/2 = h

therefore: d2-d1=2h

going back to the original problem:

Let C1 = Circumference 1
C2 = Circumference 2
D1 = Diameter 1
D2 = Diameter 2
R1 = Radius 1
R2 = Radius 2

We know that:
1. pi*D1 = C1
2. pi*D2 = C2

or

1. pi*2*R1 = C1
2. pi*2*R2 = C2

from the problem above, we can formulate that:

C1 + 1meter = C2
D2 – D1 = 2*h (based from the explanation above)
or
R2 – R1 = h

Substituting C2, D2 and R2 in terms of C1, D1 and R1 respectively:

1. pi*D1 = C1
2. pi*(D1+2*h) = C1 + 1meter

or

1. pi*2*R1 = C1
2. pi*2*(R1+h) = C1 + 1meter

Now substitute equation 1 to equation 2:

3. pi*(D1+2*h) = pi*D1 + 1meter
pi*D1 + pi*2*h = pi*D1 + 1meter
h = 1/(2*pi)
h = 0.15915 meters or 15.915 cm

or

3. pi*2*(R1+h) = pi*2*R1 + 1meter
pi*2*R1 + pi*2*h = pi*2*R1 + 1meter
pi*2*h = 1meter
h = 1/(2*pi)
h = 0.15915 meters or 15.915 cm