since the diameter of the base of the cylinder is 6 feet, then its radius is half that, or 3 feet.
![\bf \textit{volume of a cylinder}\\\\ V=\pi r^2 h~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=3\\ h=9 \end{cases}\implies V=\pi (3)^2(9)\implies V=81\pi](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20cylinder%7D%5C%5C%5C%5C%0AV%3D%5Cpi%20r%5E2%20h~~%0A%5Cbegin%7Bcases%7D%0Ar%3Dradius%5C%5C%0Ah%3Dheight%5C%5C%5B-0.5em%5D%0A%5Chrulefill%5C%5C%0Ar%3D3%5C%5C%0Ah%3D9%0A%5Cend%7Bcases%7D%5Cimplies%20V%3D%5Cpi%20%283%29%5E2%289%29%5Cimplies%20V%3D81%5Cpi)
You want to find the value of x for which the area under the curve to the left of x is 0.6. One way to do that is to create the cumulative distribution function (CDF) for the given PDF, then see where it is equal to 0.6.
Doing that, we find a = 5.
I believe the answer is A because the square root of 10 root 3 is approximately 17.32.