= Solution
Use the real mode
$$
w=A\sin(q_ny)\cos(kx-\omega t).
$$
The <kinetic energy density> and <isotropic linear-elastic energy density> are
$$
e_k=\frac12\rho w_t^2,
\qquad
e_p=\frac\mu2(w_x^2+w_y^2).
$$
Average over one period and integrate across the layer. Since $\int_0^h\sin^2(q_ny)\,dy=\int_0^h\cos^2(q_ny)\,dy=h/2$, the mean energy per unit horizontal area is
$$
\overline{\mathcal E}
=\int_0^h\overline{e_k+e_p}\,dy
=\frac{\rho A^2\omega^2h}{4}.
$$
The $x$-directed elastic-energy flux is $S_x=-\sigma_{zx}w_t=-\mu w_xw_t$. Its corresponding average is
$$
\overline{\mathcal F}_x
=\int_0^h\overline{S_x}\,dy
=\frac{\mu A^2k\omega h}{4}.
$$
Using $\omega^2=c_S^2(k^2+q_n^2)$ gives
$$
\frac{\overline{\mathcal F}_x}{\overline{\mathcal E}}
=\frac{\mu k}{\rho\omega}
=\frac{c_S^2k}{\omega}
=c_{g,n},
$$
and therefore
$$
\boxed{\overline{\mathcal F}_x=c_{g,n}\overline{\mathcal E}.}
$$
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