The radii of two circles are 8 cm and 6 cm respectively.

We have, 

Radius of circle- $\mathrm{I}, \mathrm{r}_{1}=8 \mathrm{~cm}$

Radius of circle-II, $r_{2}=6 \mathrm{~cm}$

$\therefore \quad$ Area of circle- $\mathrm{I}=\pi \mathrm{r}_{1}^{2}=\pi(8)^{2} \mathrm{~cm}^{2}$

Area of circle- $\mathrm{II}=\pi \mathrm{r}_{2}^{2}=\pi(6)^{2} \mathrm{~cm}^{2}$

Let the radius of the circle-III be R

$\therefore \quad$ Area of circle-III $=\pi R^{2}$

$\Rightarrow \pi(8)^{2}+\pi(6)^{2}=\pi R^{2}$

$\Rightarrow \pi\left(8^{2}+6^{2}\right)=\pi R^{2}$

$\Rightarrow 8^{2}+6^{2}=R^{2}$

$\Rightarrow 64+36=R^{2}$

$\Rightarrow 100=\mathrm{R}^{2}$

$\Rightarrow 10^{2}=\mathrm{R}^{2} \Rightarrow \mathrm{R}=10 \mathrm{~cm}$

Thus, the radius of the new circle $=10 \mathrm{~cm}$.