1. What is meant by average bond energy?

For $SF_6$, the average $S-F$ bond energy is the energy required to break all 6 $S-F$ bonds in 1 mole of gaseous $SF_6$ to form gaseous atoms:

$SF_6(g) \rightarrow S(g) + 6F(g)$

If the total atomization enthalpy is $\Delta H$, then

$\text{Average } S-F \text{ bond energy} = \frac{\Delta H}{6}$

2. Use enthalpy of formation data

Given:

$\Delta H_f^\circ[SF_6(g)] = -1100\ \text{kJ mol}^{-1}$ $\Delta H_f^\circ[S(g)] = 275\ \text{kJ mol}^{-1}$ $\Delta H_f^\circ[F(g)] = 80\ \text{kJ mol}^{-1}$

For the reaction

$SF_6(g) \rightarrow S(g) + 6F(g)$

by Hess's law,

$\Delta H = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})$

So,

$\Delta H = [275 + 6(80)] - [-1100]$

$= 275 + 480 + 1100$

$= 1855\ \text{kJ mol}^{-1}$

3. Find average bond energy

There are 6 identical $S-F$ bonds in $SF_6$, so

$\text{Average bond energy} = \frac{1855}{6}$

$= 309.166\ldots\ \text{kJ mol}^{-1}$

Rounded to nearest integer:

$\boxed{309\ \text{kJ mol}^{-1}}$

4. Comparison with stored answer

Stored correct answer = $309$

Our derived answer = $309$

So the answer agrees with the stored correct answer.