= Solution
For the <linear rigid rotor>, $E_\ell=\hbar^2\ell(\ell+1)/(2I)$ and the electric <electric dipole moment> interaction is $V=-E\hat d_z$. The isotropic $\ell=0$ state has no first-order shift. Its only coupled rotational state is $\ell=1,m_\ell=0$, at gap $\Delta_{\rm rot}=\hbar^2/I$, giving the <quadratic Stark shift of a polar linear rotor>
$$
\boxed{\Delta E_g=-\frac{E^2(d^2/3)}{\Delta_{\rm rot}}=-\frac{Id^2E^2}{3\hbar^2}.}
$$
Using $1\,\mathrm{kV\,cm^{-1}}=10^5\,\mathrm{V\,m^{-1}}$ and the supplied I and d gives
$$
\boxed{\Delta E_g\simeq-8.05\times10^{-27}\,\mathrm J,\qquad \Delta E_g/h\simeq-12.2\,\mathrm{MHz}.}
$$
The negative sign is the expected second-order <Stark effect> of a nondegenerate ground state. The mixing parameter $dE/(\sqrt3\Delta_{\rm rot})\simeq0.074$ is small, supporting the weak-field estimate. The constants h and $\hbar=h/(2\pi)$ use their defined SI values; no molecular input beyond the supplied data is needed.
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