Let ABCD be our trapezoid.  

All isosceles trapezoids are cyclic, and thus obey AB*CD + BC*AD = AC*BD.  

Because we are given that the trapezoid is tangential, then it must also obey AB+CD = BC+AD.  

Finally note that the triangle composed of the shorter base, lateral side and diagonal, that there is an obtuse angle opposite the diagonal which means the diagonal is longer than the shorter base and the lateral side.

Let AB and CD be the two bases, with AB as the shorter base and CD as the longer base.  Then BC and AD are the congruent lateral sides.  

Let BC=AD=n.  Then AB+CD = 2n from being tangential. 

But if the two base lengths and the lateral length are to be all from a set of four consecutive whole numbers then AB and CD must be the numbers adjacent to n, then AB=n-1 and CD=n+1.  This leaves the diagonals AC=BD=n+2.

Now we can substitute into the equation for being cyclic yielding (n-1)*(n+1) + n^2 = (n+2)^2.  This simplifies to n^2 - 4n - 5 = 0; which has two roots -1 and 5.  -1 is discarded leaving us n=5.

Then the trapezoid has bases of 4 and 6, lateral sides of 5 and diagonals of 7.  

I'll note at this point I have proven, by construction, this is the only trapezoid satisfying the problem statement (which Charlie's program fails to do).

So then just the circumradius and inradius.

Then inradius = (1/2) * sqrt[product of bases] = (1/2)*sqrt(4*6) = sqrt(6)

And circumradius = (1/4)*(sum of bases) * sqrt[1+(lateral side)^2/(product of bases)] = (1/4)*(4+6)*sqrt[1 + 5^2/(4*6)] = (5/2) * sqrt(49/24) = 35/(4*sqrt(6))

Then circumradius/inradius = (35/(4*sqrt(6))) / sqrt(6) = 35/24.