= Solution
Applying the stated long-wave expression formally in the intended high-viscosity regime, the numerator is dominated by $2\lambda\kappa^2/3$ and the displayed denominator by $\lambda$. It gives
$$
\boxed{s\sim\frac{\Delta\rho gh_0}{3\mu}\kappa^2.}
$$
The mechanism is inner transverse shear: the outer fluids almost immobilize the interfaces tangentially, so the middle layer has a pressure-driven parabolic velocity instead of the plug-like extensional velocity of part (b). Indeed, with nearly stationary faces at $z=\pm h$, its integrated flux is $-2h^3p_x/(3\mu)$. With $p_x=\Delta\rho gh_x$, <conservation of mass> gives precisely the displayed decay rate. The leading <viscous dissipation> is then in the middle layer, even though its neighbours have much greater <dynamic viscosity>.
There is a nonuniformity in taking this conclusion literally for every $\lambda\gg\kappa^{-1}$. A long-wave remainder written for fixed $\lambda$ cannot be assumed uniform as $\lambda$ diverges. Here this matters physically because sufficiently viscous outer fluids also resist normal interface motion. The distinction can be resolved directly, without an unspecified remainder, from the <Stokes equation>. For a symmetric thickness mode, the middle vertical velocity has the form
$$
w(z)=A\sinh(kz)+B(kz)\cosh(kz),\qquad u=\frac i{k}w_z.
$$
It is odd in $z$, and the middle horizontal velocity is even. At $z=h_0$, the exact outer traction map is $\sigma_{xz}^+=-2\lambda\mu k u$ and $\sigma_{zz}^+=-2\lambda\mu k w$. Tangential continuity and the normal hydrostatic jump therefore give
$$
w_{zz}+k^2w=-2\lambda k w_z,\qquad
-\frac\mu{k^2}w_{zzz}+3\mu w_z+2\lambda\mu k w=-\Delta\rho g\eta.
$$
The latter uses $p=\mu(w_{zzz}-k^2w_z)/k^2$, obtained from horizontal force balance. The kinematic condition is $w(h_0)=-s\eta$. Writing $S=\sinh\kappa$ and $C=\cosh\kappa$, tangential continuity fixes
$$
A(S+\lambda C)+B[S+\kappa C+\lambda(C+\kappa S)]=0.
$$
Eliminating $A,B$ in the remaining two equations gives the \b[exact symmetric-layer rate]
$$
s=\frac{\Delta\rho gh_0}{2\mu\kappa}
\frac{S^2+\lambda(SC-\kappa)}{\lambda\cosh(2\kappa)+(SC+\kappa)+\lambda^2(SC-\kappa)}.
$$
For $\kappa\ll1$ and $\lambda\kappa\gg1$, this proves the <uniform high-viscosity limit of viscous-layer relaxation>
$$
\boxed{s\sim\frac{\Delta\rho gh_0}{2\mu}
\frac{(2/3)\kappa^2}{1+(2/3)\lambda\kappa^3}.}
$$
Thus the intended inner-shear answer is valid in the distinguished range $\kappa^{-1}\ll\lambda\ll\kappa^{-3}$. If $\lambda\kappa^3\gg1$, the main resistance is instead outer normal motion and
$$
s\sim\frac{\Delta\rho g}{2\lambda\mu k}.
$$
Both ranges satisfy the printed $\lambda\gg\kappa^{-1}\gg1$. For example $\lambda=\kappa^{-4}$ lies in the latter range and has $s\sim[\Delta\rho gh_0/(2\mu)]\kappa^3$, rather than a rate proportional to $\kappa^2$. The exact denominator contains a $\lambda^2\kappa^3$ contribution hidden by a fixed-parameter remainder. \b[The inner-shear limit is the intended result, but an additional upper restriction on $\lambda$ is needed to make it a uniform conclusion.]
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