Let D be the diameter in cm. The number of shot in the tray is approximately 100/D^2. The equation is not accurate unless D<<10 cm. With a hexagonal arrangement of spheres, you can get about 119/D^2 speheres in. It depends upon how you arrange the spheres and how much loose space remains at the edges.
The same number fills the cylinder up to 20 cm^3. Calculate the number of spheres of diameter D that can occupy that volume with hexagonal close packing. According to a theorem of Gauss, 74% of the volume can be occupied by solid spheres, whose volume is pi*D^3/6, so the number of spheres also equals
0.74*20/(pi*D^3/6) = 28.3/D^3
28.3/D^3 = 100/D^2 ; therefore
D = 0.28 cm
You can complete the other questions with that information.
Spherical lead shot are poured into a square tray, 10 centimeters on a side, until they completely cover the bottom. The shot are poured from the tray into a graduated cylinder, which they fill to the 20-cm3 mark. What is the diameter of a single shot? How many shot were in the tray? If the 20 cm3 of shot had a mass of 130 grams, what would be the mass of a single shot?
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