= Solution
A nonzero <superconducting energy gap> suppresses the population of thermally excited <quasiparticles>. When $k_BT\ll\Delta$, their <Boltzmann factor> produces an approximately activated electronic <heat capacity>,
$$
C_e(T)\simeq A(T)\exp\left(-\frac{\Delta}{k_BT}\right),
$$
where the prefactor $A(T)$ varies algebraically rather than exponentially. For the rough estimate requested, neglect its change between the two quoted <temperatures>. Taking a ratio eliminates its unknown magnitude:
$$
\frac{C_e(3\,\mathrm K)}{C_e(2\,\mathrm K)}\simeq14
=\exp\left[\frac{\Delta}{k_B}\left(\frac1{2\,\mathrm K}-\frac1{3\,\mathrm K}\right)\right].
$$
Therefore
$$
\boxed{\frac{\Delta}{k_B}\simeq6\ln14\,\mathrm K=15.8\,\mathrm K\approx16\,\mathrm K.}
$$
The corresponding excitation <energy> is $\Delta\simeq2.19\times10^{-22}\,\mathrm J\simeq1.36\,\mathrm{meV}$. The units $\mathrm{mJ\,mol^{-1}\,K^{-1}}$ of the two heat-capacity readings cancel in their ratio.
This is a <heat-capacity activation estimate>, not an exact determination of a gap from two data points. More generally, if $C_e=AT^p e^{-\Delta/(k_BT)}$, the same measurements give
$$
\frac{\Delta}{k_B}=6\left[\ln14-p\ln\frac32\right]\mathrm K.
$$
For example, the low-temperature <BCS theory> form for a fully gapped isotropic <superconductor> has $p=-3/2$ and would give about $19.5\,\mathrm K$. The stated approximate data and constant-gap assumption support the simpler $16\,\mathrm K$ estimate; they do not specify a detailed <density of states> or heat-capacity prefactor. Here $\Delta$ means the minimum <quasiparticle> excitation <energy>, rather than automatically twice that value.
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