= Solution
One mole of the overall discharge reaction transfers one mole of electrons. Count \b[both active electrode reactants], including the graphite host:
$$
M_{\rm active}=M_{\mathrm{LiC_6}}+M_{\mathrm{CoO_2}}
=6.94+6(12.01)+58.93+2(16.00)=169.93\,\mathrm{g\,mol^{-1}}.
$$
Using the preceding constant-voltage approximation,
$$
W=FE\simeq(96500)(4.2)=405300\,\mathrm{J\,mol^{-1}}
=112.58\,\mathrm{Wh\,mol^{-1}},
$$
so the ideal active-material <specific energy> is
$$
\boxed{\varepsilon_{\rm sp}\simeq\frac{112.58}{0.16993}=663\,\mathrm{Wh\,kg^{-1}}.}
$$
This assumes complete one-electron utilization and the proxy $4.2\,\mathrm V$ throughout discharge. It is not a packaged-cell value or a cathode-only value. A varying equilibrium voltage requires integrating $\int V\,dQ$; using a maximum voltage as a constant average overestimates the delivered energy.
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