The radii of the circular ends of a frustum of height 6 cm are 14 cm and 6 cm, respectively.

We have,

Height of the frustum, $h=6 \mathrm{~cm}$,

Radii of the circular ends, $R=14 \mathrm{~cm}$ and $r=6 \mathrm{~cm}$

Let the slant height of the frustum be $l$.

Now,

$l=\sqrt{(R-r)^{2}+h^{2}}$

$=\sqrt{(14-6)^{2}+6^{2}}$

$=\sqrt{8^{2}+6^{2}}$

$=\sqrt{64+36}$

$=\sqrt{100}$

$=10 \mathrm{~cm}$

So, the slant height of the frustum is 10 cm.