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

We need to find the energy for the formation of NaBr ionic solid from its gaseous ions. The relevant energy terms are the ionization enthalpy of Na, the electron gain enthalpy of Br, and the lattice enthalpy of NaBr.

The formation of NaBr from gaseous atoms involves: $$Na_{(g)} \to Na^+_{(g)} + e^-$$ with ionization enthalpy $$= +495.8$$ kJ mol$$^{-1}$$, and $$Br_{(g)} + e^- \to Br^-_{(g)}$$ with electron gain enthalpy $$= -325.0$$ kJ mol$$^{-1}$$, and $$Na^+_{(g)} + Br^-_{(g)} \to NaBr_{(s)}$$ with lattice enthalpy $$= -728.4$$ kJ mol$$^{-1}$$.

The overall energy for the formation of NaBr ionic solid from gaseous Na and Br atoms is the sum of these three contributions: $$\Delta H = 495.8 + (-325.0) + (-728.4) = 495.8 - 325.0 - 728.4 = -557.6 \text{ kJ mol}^{-1}$$

The question asks for the answer in the form $$(-)\underline{\hspace{1cm}} \times 10^{-1}$$ kJ mol$$^{-1}$$. Since $$-557.6 = (-) 5576 \times 10^{-1}$$, the answer is $$\mathbf{5576}$$.