Too easy geometry!

We recognize the right triangle with sides of $6$ and $8$ as a scaled up version of the 3-4-5-triangle. Therefore the third side of the triangle is $10$ units. If the radii of the circles are $a$ , $b$ and $c$ , we can set up the equations

$a+b=6$

$a+c=8$

$b+c=10$

We can solve this linear system of equations to get

$a=2$

$b=4$

$c=6$

The circumference of a circle with radius $r$ is given by $2 \pi r$ , so the circumferences of the three circles are

$2 \pi \cdot a = 4\pi$

$2 \pi \cdot b = 8\pi$

$2 \pi \cdot c = 12\pi$

Their sum is

$4\pi + 8\pi + 12\pi = \boxed{24\pi}$