= Solution
The largest applied field gives
$$
\frac{\mu E_{\rm field}}{hc}=\frac{(1.6)(3.33564\times10^{-30})(10^4)}{(6.62607015\times10^{-34})(2.99792458\times10^{10})}
\simeq2.69\times10^{-3}\ \mathrm{cm^{-1}}.
$$
For parallel <polarization>, adjacent allowed M values in a $J,K$ component are separated by
$$
\Delta_M\widetilde\nu=\frac{2(2.69\times10^{-3})|K|}{J(J+1)(J+2)}.
$$
Already at $K=1$ this is \b[$2.24\times10^{-4}\ \mathrm{cm^{-1}}$ for $J=2$] and \b[$8.96\times10^{-5}\ \mathrm{cm^{-1}}$ for $J=3$]. Both exceed the stated $2\times10^{-5}\ \mathrm{cm^{-1}}$ resolution, and larger $|K|$ gives larger first-order spacings. \b[The spectrometer should resolve suitable Stark components], provided their physical linewidths and signal-to-noise permit it. This predicts resolvable splittings rather than incorrectly comparing the maximum level shift directly with instrumental resolution.
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