Baseball, neutrons, and protons are all perfect spheres.
Circumference of the baseball,
.
mass of the proton or neutron, 
diameter of the proton or neutron, 
Mass per unit volume is the definition of a material's density. The density equation is

Here, M is the mass of the material, V is the volume, and
is the density of the material.
We know that formula for the volume of a sphere having diameter d is 
You may obtain this result by plugging this number into the proton/neutron density equation:

By replacing the variables in the aforementioned equation with their values, you will obtain

The formula for calculating a circle's circumference with diameter d is 
Using this, you get the diameter of the baseball:

The volume of the baseball may be calculated using the formula for the volume of a sphere:

The mass of the baseball is determined by the concept of density if you suppose that its density is equal to that of a proton or neutron.

Here,
is the density of the proton/neutron.
By replacing the variables in the aforementioned equation with their values, you will obtain

Therefore, if the density of the baseball was equal to that of the neutron/proton, then its mass would be 
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