Assuming the experimental mass of \({}^{12}_{6}C\) as \(12\text{ u}\), the mass defect of \({}^{12}_{6}C\) atom is _______ \(\text{MeV}/c^2\).
(Mass of proton \(= 1.00727\text{ u}\), mass of neutron \(= 1.00866\text{ u}\), \(1\text{ u} = 931.5\text{ MeV}/c^2\) and \(c\) is the speed of the light in vacuum).

The mass of \(6\) protons is \(6 \times 1.00727\text{ u} = 6.04362\text{ u}\).

The mass of \(6\) neutrons is \(6 \times 1.00866\text{ u} = 6.05196\text{ u}\).

Total mass of the constituent nucleons is \(6.04362\text{ u} + 6.05196\text{ u} = 12.09558\text{ u}\).

The mass defect \(\Delta m\) is the difference between the total mass of the nucleons and the experimental mass of the atom (ignoring electron mass as per the given data):

\(\Delta m = 12.09558\text{ u} - 12\text{ u} = 0.09558\text{ u}\)

Converting the mass defect into \(\text{MeV}/c^2\):

\(\Delta m = 0.09558 \times 931.5\text{ MeV}/c^2 = 89.03277\text{ MeV}/c^2 \approx 89.03\text{ MeV}/c^2\)

Answer: \(89.03\)