Because the sums of both the top and bottom numbers (the mass number and atomic numbers) of each side must be equal, we know that in A the atomic number must be 2 because 88+2=90 and the mass number must be 4 so 228+4=232. Thus, because the atomic number is 2, you know this is helium and the nuclide is 4/2 He. This is a classic example of alpha decay.
In letter B, the same concept can be applied, except the helium nuclide is given. The atomic number must be 86 so 86+2=88 and the mass number must be 220 so 220+4=224. Because 86 is the atomic number of radon, the nuclide is 220/86 Rn.
For letter C, the gamma emission can be ignored as it has no mass. The mass number must be 4+237=241 and the atomic number must be 2+93=95. Because 95 is the atomic number of americium, the nuclide is 241/95 Am.
