Answer
$4.678\;\rm eV$
Work Step by Step
We know that the two electrons spend 33$\%$ of the time between the two protons (the two hydrogen atoms since each atom contains one proton), thus the net charge there between the two protons, as we see below, is
$$q_e=2\times 0.33e$$
$$q_e=0.66e\tag 1$$
The potential energy of the charge $q_e$ midway between the two protons is given by
$$PE=2PE_{q_e\rightarrow q_p}=2\dfrac{-kq_eq_p}{r}=2\dfrac{-kq_eq_p}{\frac{1}{2}d}$$
where $2$ in front of $PE_{q_e\rightarrow q_p}$ is due to the potential energy due to each proton on the negative charge $q_e$, and $r$ is half the separation distance between the two protons, as we see in the figure below.
$$PE_e=\dfrac{-4kq_eq_p}{ d} $$
Plugging from (1);
$$PE_{e}=\dfrac{-4kq_eq_p}{ d} \tag 2$$
And the potential energy between the two protons is given by
$$PE_p=\dfrac{kq_p^2}{d}\tag 3$$
We know that the initial energy of each hydrogen atom in the ground state is $E_1=-13.6\rm \;eV$, and hence when the bond between these two atoms breaks each one will have this ground state of -13.6 eV.
Thus, the binding energy is given by
$$E_{binding}=2E_1-\left[PE_e+PE_p\right]$$
Plugging from (2) and (3);
$$E_{binding}=2E_1-\left[\dfrac{-4kq_eq_p}{ d}+\dfrac{kq_p^2}{d}\right]$$
$$E_{binding}=2E_1-\left[\dfrac{-4kq_eq_p +kq_p ^2}{ d} \right]$$
Plugging from (1), and recalling that $q_p=e$;
$$E_{binding}=2E_1-\left[\dfrac{-4k(0.66e^2) +ke^2}{ d} \right]$$
$$E_{binding}=2E_1-\left[\dfrac{-1.64ke^2 }{ d} \right]$$
Plugging the known;
$$E_{binding}=(2\times-13.6\times 1.6\times10^{-19})-\left[\dfrac{-1.64\times 8.99\times10^9\times (1.6\times10^{-19})^2 }{ 0.074\times10^{-9}} \right]$$
$$E_{binding}=\color{red}{\bf 7.48489\times10^{-19}}\;\rm J\approx \color{red}{\bf 4.678}\;eV$$
