I suggest you simply google "volume of frustrum" for the volume formula.
Then subtract the volume of the cylinder inside (google "volume of cylinder", but you should know that!)
The first thing I would do is draw an outline of the cone on graph paper (or at least imagine that being done). The top edge runs from (0, 5.3) to (46, 7.7). Since that is a straight line, its equation is linear: y= ax+ b for some numbers a and b. Setting x= 0, y= 5.3 we have 5.3= a(0)+ b= b. Setting x= 46, y= 7.7, 7.7= a(46)+ b= 46a+ 5.3 so 46a= 7.7- 5.3= 2.4. a= 2.4/46= 0.05217 (rounded to 4 significant figures). In this graph, the top edge is given by y= 0.05217x+ 5.3 where y= 0 is the axis of the cone. The cone itself is that line rotated around the x-axis. Now, imagine slicing that cone perpendicular to the axis. At each x we get disks with radius 0.05217x+ 5.3 so area $\displaystyle \pi (0.5217x+ 5.3)^2$. Imagine cutting the cone into a large number of those disks with thickness "$\displaystyle \Delta x$". Those aren't actually disks since the radius on one side is slightly less than the radius on the other side (0.5217x+ 5.3 on one side and 0.5217(x+ \Delta x)+ 5.3 on the other. But if we ignore that slight difference the volume of each "disk" is $\displaystyle \pi(0.5217x+ 5.3)^2\Delta x$ and the volume of the entire cone is approximately $\displaystyle \sum\pi(0.5217x+ 5.3)^2$. Taking the limit as the number of disks goes to infinity and the thickness of each disk goes to 0, that volume becomes the integral $\displaystyle \pi \int_0^{46} (0.5217x+ 5.3)^2 dx$. That will give the same thig as the "volume of frustrum formula" Debsta refers to. Then, as Debsta says, you need to subtract the volume of the central core.
You could do the same thing, dividing the length of the core into small thicknesses to get an integral but, since the radius of a cylinder is a constant, here 2.8 cm, that volume is just the "cross section area", $\displaystyle \pi (2.8)^2= 7.84\pi$, times the length, 46 cm, so $\displaystyle \pi(2.8)^2(46)= 1133$ cubic centimeters.
(Why was this posted under "Math Software"?)