According to the explanation given in the previous session, in this case we have an element with the following configuration: 112/49 X (I will call it X)
A = n + p, atomic mass is the sum of protons and neutrons
Z = p = e, atomic number is equal to number of protons and electrons (in a NEUTRAL atom)
Z = 49, therefore we have 49 protons and 49 electrons
A = 112
112 = 49 + n
n = 112 - 49
n = 63 neutrons
Therefore the best option is letter A, 49 protons, 63 neutrons and 49 electrons