Solution:
Since $\triangle ABC$ is a right triangle, its circumradius is just half of its hypotenuse. By Pythagorean theorem, $AC = \sqrt{6^{2} + 8^{2}} = 10$ and so $R = \frac{AC}{2} = 10 / 2 = 5$.

Now, we can compute the inradius by using the following identity:
$$\left(r \times \frac{AB + BC + AC}{2} = \operatorname{Area}(\triangle ABC)\right)$$
Since $\triangle ABC$ is a right triangle, its area is just $(AB \times BC) / 2 = 48 / 2 = 24$. So that
$$r = \frac{24}{(6 + 8 + 10) / 2} = 2$$
Thus, $R + r = 5 + 2 = 7$.