= Solution
Use the same molar amount for both <thermodynamic phases>. A hexagonal cell built from two basal lattice vectors at120degrees has volume $(\sqrt3/2)a^2c$, not the volume of a three-times-larger full hexagonal prism. With four atoms per stated <graphite> cell,
$$
V_{m,g}=\frac{N_A}{4}\frac{\sqrt3}{2}
(0.246\times10^{-9})^2(0.671\times10^{-9})
\simeq5.2944\times10^{-6}\ {\rm m^3\,mol^{-1}}.
$$
<Diamond> has eight atoms per conventional cubic cell, so
$$
V_{m,d}=\frac{N_A}{8}(0.357\times10^{-9})^3
\simeq3.4250\times10^{-6}\ {\rm m^3\,mol^{-1}}.
$$
For <graphite> transforming to <diamond>,
$$
\Delta V_m\simeq-1.8693\times10^{-6}\ {\rm m^3\,mol^{-1}},
\qquad
\Delta S_m=2.45-5.69=-3.24\ {\rm J\,K^{-1}\,mol^{-1}}.
$$
Therefore
$$
\boxed{\frac{\mathrm dP}{\mathrm dT}
\simeq1.73\times10^6\ {\rm Pa\,K^{-1}}
=1.73\ {\rm MPa\,K^{-1}}.}
$$
The positive slope reflects <diamond>'s smaller <entropy> and volume. Higher temperature favors the higher-entropy <graphite> <thermodynamic phase>, requiring greater pressure to compensate in favor of the smaller-volume <diamond> <thermodynamic phase>.
This estimate treats the supplied ambient molar <entropies> and volumes as applicable near coexistence and neglects their changes through <thermal expansion>, <compressibility> and <heat capacity>. It gives an approximate gradient, not the transition pressure or a complete high-pressure <thermodynamic phase> boundary; an intercept would require an additional free-energy datum.
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