$\frac{1}{2}{S}_{8\left(g\right)}+6{\mathrm{O}}_{2\left(9\right)}⟶4{\mathrm{SO}}_3;\Delta \mathrm{H}°=-1590\mathrm{KJ}$ The standard enthalpy of formation of ${\mathrm{SO}}_3$ is

$\Delta H_f=\Delta H_f$ (products) $-\Delta H_f($ reactants $)$

 Given $\Delta H_f=-1590\mathrm{KJ}$

 The reactants are in elemental form and hence the enthalpy of formation is zero

 By substituting the values in the above equation

 $-1590=4\times \Delta H_f\left({\mathrm{SO}}_3\right)-0 =4$ moles of sulphur trioxide are formed

 $H_f\left({\mathrm{SO}}_3\right)=-\frac{1590}{4}=-397.5$

Hence the correct option is (B).