Answer
$13.22\;\rm eV$
Work Step by Step
The uncertainty of the momentum of one electron in one dimension ($x$-direction) is given by
$$\Delta p_x=\dfrac{\hslash}{\Delta x}$$
and we know that the momentum of the electron is at least equal to the uncertainty of the momentum. Thus,
$$ p_x\approx \dfrac{\hslash}{\Delta x}\tag 1$$
The kinetic energy of the electron is given by
$$KE=\dfrac{p_x^2}{2m_e}$$
Plugging from (1);
$$KE=\dfrac{\left[\dfrac{\hslash}{\Delta x}\right]^2}{2m_e}$$
$$KE=\dfrac{\left[\dfrac{h}{2\pi \Delta x}\right]^2}{2m_e}$$
$$KE=\dfrac{h^2}{8\pi^2m_e\Delta x^2}$$
This is the kinetic energy of one electron, so for two electrons,
$$KE_{tot}=2\times\dfrac{h^2}{8\pi^2m_e\Delta x^2}$$
$$KE_{tot}= \dfrac{h^2}{4\pi^2m_e\Delta x^2}\tag 2$$
Therefore, the difference in kinetic energies between the in-atom electrons and the in-molecule electrons is given by
$$\Delta KE=KE_{tot, atom}-KE_{tot, molecule }$$
Plugging from (2);
$$\Delta KE=\dfrac{h^2}{4\pi^2m_e\Delta x^2_{atom}} -\dfrac{h^2}{4\pi^2m_e\Delta x^2_{molecule }}$$
$$\Delta KE=\dfrac{h^2}{4\pi^2m_e } \left[\dfrac{1}{ \Delta x^2_{atom}}-\dfrac{1}{ \Delta x^2_{molecule }}\right]$$
Plugging the known;
$$\Delta KE=\dfrac{(6.626\times 10^{-34})^2}{4\pi^2(9.11\times10^{-31}) } \left[\dfrac{1}{(0.053\times10^{-9})^2 }-\dfrac{1}{ (0.074\times10^{-9})^2}\right]$$
$$\Delta KE= \color{red}{\bf 2.11\times 10^{-18}}\;\rm J= \color{red}{\bf13.22}\;\rm eV$$