Let $a$, $b$, $c$ be the sides such that $a = b - 1$ and $c = b + 1$. Recall Heron's Formula and two other well known formulae for the area of a triangle:
$$|ABC| = \sqrt{s(s-a)(s-b)(s-c)} = \frac{abc}{4R} = rs,$$
where $r = 4$ is the inradius and $R$ the circumradius.
We obtain $s = ((b - 1) + b + (b + 1))/2 = 3b/2$ and so $|ABC| = rs = 6b$. Therefore, $36b^2 = |ABC|^2 = s(s - a)(s - b)(s - c) = 3b^2(b^2 - 4)/16$, i.e. $192 = b^2 - 4$ and so $b^2 - 1 = 195$. This gives $b^2 = 196 = 14^2$ and so $b = 14$, $a = 13$ and $c = 15$. Finally, $6b = |ABC| = \frac{abc}{4R}$ and so
$$R = \frac{abc}{24b} = \frac{b^2 - 1}{24} = \frac{195}{24} = \frac{65}{8}.$$