= Solution
The $K=0$ lines give two equations independent of $\widetilde D_{JK}$:
$$
1.50130=6\widetilde B-108\widetilde D_J,\qquad
2.00171=8\widetilde B-256\widetilde D_J.
$$
Solving yields \b[$\widetilde B\simeq0.25022\ \mathrm{cm^{-1}}$ and $\widetilde D_J\simeq2.1\times10^{-7}\ \mathrm{cm^{-1}}$]. The small distortion is obtained by subtracting close numbers, so its precision is considerably worse than that of $\widetilde B$.
For components in a fixed group,
$$
\widetilde\nu(J,0)-\widetilde\nu(J,K)=2\widetilde D_{JK}(J+1)K^2.
$$
For example the $J=2,K=2$ displacement gives $\widetilde D_{JK}=0.00008/24$, while $J=3,K=3$ gives $0.00024/72$. Both give \b[$\widetilde D_{JK}\simeq3.3\times10^{-6}\ \mathrm{cm^{-1}}$]; the smaller gaps are consistent after rounding. The pure rotational transitions have $\Delta K=0$, so the $(\widetilde A-\widetilde B)K^2$ and $\widetilde D_KK^4$ terms cancel. \b[Neither $\widetilde A$ nor $\widetilde D_K$ can be determined from these zero-field lines.]
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