Let point $P(x_1, y_1)$ be any point on the circle, therefore it satisfies the circle

${\left(x_1-3\right)}^2+{\left(y_1+2\right)}^2=5r^2....\left(i\right)$

The length of the tangent drawn from point $P(x_1,y_1)$ to the circle

${\left(x-3\right)}^2+{\left(y+2\right)}^2=r^2$ is $\sqrt{S_1}$

$=\sqrt{{\left(x_1-3\right)}^2+{\left(y_1+2\right)}^2-r^2}$

$=\sqrt{5r^2-r^2}$ [from Eq. $\left(i\right)$]

$=2r$

Given that length of tangent is $16$ units

$\Rightarrow 16=2r \Rightarrow r=8$ units

$\therefore$ The area between two circles is represented by shaded portion

So, area $=\pi \left(5r^2\right)-\pi \left(r^2\right)$ $=4\pi r^2$

$=4\pi \times 8^2$

$=256\pi$ sq. units.