Cone of tangents

Let $O$ denote the sphere center. Distance $AO=\sqrt{(5-1)^2+(10-5)^2+(16-4)^2}=\sqrt{185}$ . Dimensions of the cone can be obtained in 2D by constructing a circle of radius 5 and an apex distanced $\sqrt{185}$ from the center of the circle.

Applying Pythagorean theorem:

$AB=4 \sqrt{10}$

$\triangle ABO=10\sqrt{10} \Rightarrow BD =2 \times \triangle ABO / AO \approx 4.65$ .

$AD = \sqrt{AB^2-BD^2} \approx 11.76$

$r=BD, h=AD$ .

$V=\frac{1}{3} \pi r^2 h \approx \boxed{266.3}$