The mass of proton, neutron and helium nucleus are respectively 1.0073 u, 1.0087 u and 4.0015u. The binding energy of helium nucleus is:

We have the mass of a proton $$m_p = 1.0073$$ u, mass of a neutron $$m_n = 1.0087$$ u, and mass of a helium nucleus $$m_{He} = 4.0015$$ u. A helium nucleus ($$^4_2He$$) contains 2 protons and 2 neutrons.

The total mass of the individual nucleons is:

$$M_{nucleons} = 2m_p + 2m_n = 2(1.0073) + 2(1.0087)$$

$$= 2.0146 + 2.0174 = 4.0320 \text{ u}$$

Now the mass defect is:

$$\Delta m = M_{nucleons} - m_{He} = 4.0320 - 4.0015 = 0.0305 \text{ u}$$

Using the conversion factor $$1 \text{ u} = 931.5 \text{ MeV/c}^2$$, the binding energy is:

$$BE = \Delta m \times 931.5 \text{ MeV} = 0.0305 \times 931.5$$

$$BE = 28.41 \text{ MeV} \approx 28.4 \text{ MeV}$$

So, the answer is $$28.4$$ MeV.