= Solution
Write $\mathbf g=-g\mathbf e_z$. The vertical component of the balance gives <hydrostatic pressure>,
$$
p(x,y,z)=p_0+\rho g[h(x,y)-z].
$$
Its horizontal <gradient> is $\nabla_h p=\rho g\nabla_hh$. The horizontal balance is consequently
$$
f\mathbf e_z\mathbin\times\mathbf u=-g\nabla_hh,
\qquad
\mathbf u=\frac gf\mathbf e_z\mathbin\times\nabla_hh.
$$
Thus $\mathbf u\mathbin\cdot\nabla_hh=0$: the velocity is tangent to every <level set> of $h$, so those contours are <streamlines> of the <geostrophic flow>.
Across a vertical section joining the contours $h_0$ and $h_0+\Delta h$, an element of horizontal distance normal to the contours is $dn=dh/|\nabla_hh|$, while $|\mathbf u|=(g/f)|\nabla_hh|$. The <volumetric flow rate> is therefore
$$
Q=\int_{h_0}^{h_0+\Delta h}h\frac gf|\nabla_hh|\frac{dh}{|\nabla_hh|}
=\frac g{2f}[(h_0+\Delta h)^2-h_0^2]
\sim\boxed{\frac{gh_0\Delta h}{f}}.
$$
For <dimensional analysis>, $[g]=LT^{-2}$, $[f]=T^{-1}$ and $[h_0]=[\Delta h]=L$, so the result has dimension $L^3T^{-1}$, exactly that of a volume flux.
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