= Solution
A numerical scale estimate requires a reference ocean <density> and <latitude> not specified exactly. Taking $\rho_0\simeq1025\,\mathrm{kg\,m^{-3}}$, $g\simeq9.81\,\mathrm{m\,s^{-2}}$ and a representative midlatitude $|f|\simeq10^{-4}\,\mathrm{s^{-1}}$ gives
$$
g'\simeq9.6\times10^{-3}\,\mathrm{m\,s^{-2}},\qquad
c=\sqrt{g'h_0}\simeq2.2\,\mathrm{m\,s^{-1}},\qquad
\boxed{L_R\simeq22\,\mathrm{km}.}
$$
For a small-amplitude interface displacement of scale $L=1\,\mathrm{km}\ll L_R$, gravity drives rapid horizontal acceleration and radiates most of the displacement in <Inertia-gravity wave>[inertia-gravity waves]. The gravity adjustment time is $L/c\sim8$ minutes, much shorter than $1/|f|\sim2.8$ hours. A weak balanced remnant remains in an unbounded inviscid adjustment problem; <energy> is carried away by waves, not dissipated merely by <geostrophic adjustment>.
For $L=100\,\mathrm{km}\gg L_R$, rotational adjustment produces a largely retained interface displacement with a <Geostrophic flow>[geostrophic current]. Wave emission and <Inertial oscillation>[inertial oscillations] accompany the adjustment, but the balanced structure is much more important. To quantify that distinction under an initially resting, linear single-Fourier-component assumption, potential-vorticity conservation gives
$$
\frac{\eta_{\rm balanced}}{\eta_{\rm initial}}
=\frac{1}{1+(KL_R)^2},\qquad K\sim1/L.
$$
This is about $0.002$ for the kilometre scale and $0.95$ for the hundred-kilometre scale. These fractions use $L$ as an inverse <wavenumber> scale, not a full <wavelength>, and are illustrative rather than unique predictions for an unspecified disturbance or bounded basin.
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