I drew a picture of your truncated cone and extended it to a vertex P.
In the diagram the distance PQ is x mm. Using similar triangles
(600 + x)/1475 = x/447
Solving for x I found that x = 260.89 mm.
Using Pythagoras theorem with the triangles PQT and PRS I found |PT| = 517.56 mm and |PS| = 1707.85 mm. Thus the plate can be contructed from a disk of radius
|PS| = 1707.85 mm by removing a cisk of radius |PT| = 517.56 mm.
Using the same method as in the answer to a previous problem I found the angle VPS to be 49.09 degrees.