Answer
28.32 MeV.
Work Step by Step
The required energy provides the increase in rest energy.
$$E=[(2m_{p+e}+2m_n)-m_{He}]c^2$$
$$E=[(2(1.00783u)+2(1.00867u))-4.00260u]c^2(\frac{931.5MeV/c^2}{u})$$
$$=28.32MeV$$
This is the total binding energy of the helium nucleus.