 Solution

The Fermi temperature isInserting it into the question's quoted estimate givesAs explained in (d), that count is per mole of conduction electrons. Using the stipulated six electrons per tungsten atom gives $1.8025\times 10^{-3}\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ per mole of tungsten in the same rough model. The accurate free-electron Sommerfeld expansion coefficient would instead give $4.4474\times 10^{-3}\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ per mole of tungsten. These are different approximations and counting conventions, not three interchangeable exact answers.