Equation of the ellipse is $\frac{x^2}{25}+\frac{y^2}{9}=1$
Any point on the ellipse is $\left(5\cos \theta ,3\sin \theta \right)$ at which the equation of the tangent is
$\frac{x}{5}\cos \theta +\frac{y}{3}\sin \theta =1$
If it is also a tangent to the circle $x^2+y^2=16$, then its distance  from the origin is equal to the radius, 
$\Rightarrow \frac{1}{\sqrt{\left(\frac{\mathrm{c}\mathrm{o}{\mathrm{s}}^2\theta }{25}+\frac{\mathrm{s}\mathrm{i}{\mathrm{n}}^2\theta }{9}\right)}}=4$
$\Rightarrow \frac{{\cos }^2\theta }{25}+\frac{1-{\cos }^2\theta }{9}=\frac{1}{16}$
$\Rightarrow \cos \theta =\frac{5\sqrt{7}}{16}$
If $l$ be the length of the tangent then $l^2=S^{\text{'}}$
or $l^2=25\mathrm{c}\mathrm{o}{\mathrm{s}}^2\theta +9\mathrm{s}\mathrm{i}{\mathrm{n}}^2\theta -16=16\mathrm{c}\mathrm{o}{\mathrm{s}}^2\theta -7$
$=16.\frac{25.7}{{\left(16\right)}^2}-7=\frac{175-112}{16}=\frac{63}{16}$
$\Rightarrow l=\frac{3}{4}\sqrt{7}$ units