Answer
$0.094$ or, $9.4 \%$
Work Step by Step
The area of the unshaded region is equal to $6(\dfrac{1}{2}\times 10.4 \times 9)- \pi (9)^2=(280.8-81\pi)m^2$
Now, the probability that the point lies in the unshaded region is the ratio of area of the shaded region to the total area:
P(point lies in unshaded region)$=\dfrac{280.8-81 \pi}{280.8}=0.094$ or, $9.4 \%$
