For the transition ${\mathrm{C}}_{\text{(diamond) }}\to {\mathrm{C}}_{\text{(graphite) }}$; $\Delta \mathrm{H}=-1.5\mathrm{KJ}$ It follows that

$\Delta {\mathrm{H}}_{\text{transition }}$ of $\mathrm{C}$ (graphite) $\to \mathrm{C}$ (diamond) is $1.9\mathrm{ }\mathrm{k}\mathrm{J}/\mathrm{mol}$ at $25°\mathrm{C}$ so its an endothermic process.

The internal energy of a diamond is more than graphite so it has higher combustion energy.

Also, as the internal energy of graphite is less than diamond so $\mathrm{C}$ (graphite) is more thermodynamically stable than $\mathrm{C}$ (diamond) at $25°\mathrm{C}$.

$\Delta {\mathrm{G}}_{\text{transition }}$ of $\mathrm{C}$ (diamond) $\to \mathrm{C}$ (graphite) is -ve as for this process $\Delta \mathrm{H}$ is negative and $\Delta \mathrm{S}$ is positive so $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T}\Delta \mathrm{S}$ will always negative. During the transformation of diamond to graphite, energy is released. This implies that graphite has lower energy content and, hence, is more stable than diamond.

Hence the correct answer is (A) Graphite is stabler than diamond.