The ionization enthalpy of ${\mathrm{Na}}^{+}$ formation from $\mathrm{Na}\left(\mathrm{g}\right)$ is $495.8 \mathrm{kJ}{ \mathrm{mol}}^{-1}$, while the electron gain enthalpy of $\mathrm{Br}$ is $-325.0 \mathrm{kJ}{ \mathrm{mol}}^{-1}$. Given the lattice enthalpy of $\mathrm{NaBr}$ is $-728.4 \mathrm{kJ} mo{l}^{-1}$. The energy for the formation of $\mathrm{NaBr}$ ionic solid is $\left(-\right)$______$\times {10}^{-1} \mathrm{kJ}{ \mathrm{mol}}^{-1}$.

The chemical reactions provided in the question can be written as:

$\mathrm{Na}\left(\mathrm{s}\right)\to {\mathrm{Na}}^{+}\left(\mathrm{g}\right)+{\mathrm{e}}^{-} \Delta \mathrm{H}=495.8 \mathrm{kJ}{ \mathrm{mol}}^{-1}$

$\frac{1}{2}{\mathrm{Br}}_2\left(\mathrm{I}\right)+{\mathrm{e}}^{-}\to {\mathrm{Br}}^{-}\left(\mathrm{g}\right) \Delta \mathrm{H}=-325.0 \mathrm{kJ}{ \mathrm{mol}}^{-1}$

${\mathrm{Na}}^{+}\left(\mathrm{g}\right)+{\mathrm{Br}}^{-}\left(\mathrm{g}\right)\to \mathrm{NaBr}\left(\mathrm{s}\right) \Delta \mathrm{H}=-728.4 \mathrm{kJ}{ \mathrm{mol}}^{-1}$

Adding these three chemical equations, we get:

$\mathrm{Na}\left(\mathrm{s}\right)+\frac{1}{2}{\mathrm{Br}}_2\left(\mathrm{I}\right)\to \mathrm{NaBr}\left(\mathrm{s}\right) \Delta {\mathrm{H}}_{\mathrm{f}}=?$

$\Delta {\mathrm{H}}_{\mathrm{f}}=495.8-325.0-728.4=-557.6 \mathrm{kJ}{ \mathrm{mol}}^{-1}$

$\Delta {\mathrm{H}}_{\mathrm{f}}=-5576\times {10}^{-1} \mathrm{kJ} {\mathrm{mol}}^{-1}$

Hence, the required answer is 5576.