In the right-angled triangle \(\triangle ABC\): \[ AC^2 = AB^2 + BC^2 \Rightarrow AC^2 = 6^2 + 8^2 = 36 + 64 = 100 \Rightarrow AC = \sqrt{100} = 10 \] In a right-angled triangle, the circumradius is half of the hypotenuse. \[ \therefore\ \text{Circumradius of } \triangle ABC = \frac{10}{2} \text{ cm} = 5 cm \] (Answer)