Radius of the conical vessel, $R=AC=6cm$

Height of the conical vessel, $h=OC=8cm$

Radius of the sphere, $PD=PC=r$

$\therefore PC=PD=rAC=AD=6\text{cm}$

[Since, lengths of two tangents from an external point to a circle are equal] 
$△OCA\&△OPD$ are right triangle.
[$\because$ Tangent and radius  are perpendicular to each other] 
$OA=\sqrt{O{C}^2+A{C}^2}=\sqrt{8^2+6^2}=\sqrt{100}=10\text{cm}O{P}^2=O{D}^2+P{D}^2$
$OD=OA-AD=10-6=4\text{cm}OP=OC-PC=8-r(8-r{)}^2=4^2+r^{2}64-16r+r^2=16+r^{2}16r=48\Rightarrow r=3\text{cm}$. 
Volume of water overflown =  Volume
                                                 of sphere
                                 $=\frac{4}{3}\pi r^3=\frac{4}{3}\pi \times (3{)}^3=36\pi {\text{cm}}^3$
Original volume of water = volume of
                                               cone 
                                       $=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi \times 6^2\times 8=96\pi {\text{cm}}^3$ 

$\therefore$  Fraction of water overflown $=\frac{\text{Volume of water overflown}}{\text{Original volume of water}}=\frac{36\pi }{96\pi }=\frac{3}{8}=0.375$ 
$\therefore$  Fraction of water overflown is $0.375$