The side of this regular heptagon is 1 (given).  

Let O be the center of the heptagon ABCDEFG.

Let the radius of the circumscribed circle around the heptagon be r


Its central angle is a =  = 51.4286 degrees.


For the radius r we have this equation

    r*sin(a/2) = 1/2,  which gives  r =  =  = 1.1524.


Now consider triangle OCF.  It is isosceles triangle.

Its lateral sides OC and OF have the length r, and they conclude the angle COF of 3a =  = 154.2857 degrees.


So, the length of CF is (use the cosine law for triangle OCF)

    |CF| =  =  =  = 2.247.



Let O' be the center of the circle, which touches  CD at C  and  touches EF at F.

Let R be the radius of this circle, which the problem asks to determine.


The angle at O' between perpendiculars to CD at C  and  to EF at F is 2a.


Now apply the cosine law to triangle  O'CF

    R^2 + R^2 - 2R*R*cos(2a) = |CF|^2


and find

    R =  =  =  = 1.4354.    ANSWER

Since no one has drawn the figure, and also has not given an exact value
for the radius, I thought I would do so, with an exact solution in terms 
of trigonometric values.

We draw perpendiculars to DC at C and to EF at F, and they must intersect
at the center of the circle.  We also draw FC to make an isosceles trapezoid
from which we can get the measurements for the angles, in particular angle CFE,
which turns out to be  radians.  

The sum of the interior angles of a polygon with n-sides = 
So each interior angle of the regular heptagon is , so angles E and D are  each.

Isosceles trapezoid FEDC has sum of interior angles  so angles EFC and
FCD are each 
 



Next we'll draw in 3 perpendiculars to FC, namely EH, DJ, and OI.
Notice that angles EFH and FOI have equal measures because they are both
complements of the same angle IFO. Therefore angle FOI also measures .



Since each side of the heptagon is 1, 
(a) the hypotenuse EF of right triangle EFH is 1 and 
(b)  because HJ=ED=1 and HI is 1/2 of HJ
(c) 

So from right triangle FIO, we can now find the desired radius FO

 and 

So the exact answer for the radius is

 which can also be written as

  <--EXACT SOLUTION!

That's approximately 1.436997393, which approximately agrees with Ikleyn's,
and is even closer to Greenestamps', approximate solution. 

Edwin