Given 
  $$\begin{gathered} {\mathrm{E}}_{\mathrm{H}-\mathrm{H}} = 104.2 \mathrm{kcal} {\mathrm{mol}}^{-1} \\ {\mathrm{E}}_{\mathrm{F}-\mathrm{F}} = 36.6 \mathrm{kcal} {\mathrm{mol}}^{-1} \\ {\mathrm{E}}_{\mathrm{H}-\mathrm{F}} = 134.6 \mathrm{kcal} {\mathrm{mol}}^{-1} \end{gathered}$$
The electronegativity of fluroine is :

$$\begin{gathered} {\mathrm{X}}_{\mathrm{F}}-{\mathrm{X}}_{\mathrm{H}} = 0.208{[{\mathrm{E}}_{\mathrm{F}-\mathrm{H}}-\frac{1}{2}({\mathrm{E}}_{\mathrm{H}-\mathrm{H}}+{\mathrm{E}}_{\mathrm{F}-\mathrm{F}}\left)\right]}^{\frac{1}{2}} \\ = 0.208{[134.6-\frac{1}{2}(104.2+36.6\left)\right]}^{\frac{1}{2}} \\ = 0.208{[134.6-70.4]}^{\frac{1}{2}} = 1.664 \\ {\mathrm{X}}_{\mathrm{F}} = 2.1 +1.664 = 3.764 \approx 4.0 \end{gathered}$$