= Solution
Here $\beta=p_{\rm rad}/p_{\rm gas}$ is the <radiation-to-gas pressure ratio>, not <plasma beta>. Write the <opacity> as $\kappa_{\rm op}$ to distinguish it from the <radial epicyclic frequency>. The <radiation pressure> is $p_{\rm rad}=4\sigma T^4/(3c)$, so <radiative diffusion> can be written as
$$
F_z=-\frac{c}{\kappa_{\rm op}\rho}\frac{dp_{\rm rad}}{dz}.
$$
Since $\beta$ is independent of height, $p_{\rm rad}=\beta p/(1+\beta)$. Substitution of <hydrostatic equilibrium>, $p'=-\rho\Omega^2z$, gives
$$
F_z=\frac\beta{1+\beta}\frac{c\Omega^2}{\kappa_{\rm op}}z.
$$
Differentiating and using the viscous heating equation, together with $r\Omega'=-q\Omega$, yields
$$
\rho\nu q^2\Omega^2=\frac\beta{1+\beta}\frac{c\Omega^2}{\kappa_{\rm op}},\qquad
\boxed{\rho\nu=\frac\beta{1+\beta}\frac{c}{q^2\kappa_{\rm op}}}.
$$
Thus the effective <dynamic viscosity> $\rho\nu$ is independent of height. If the surfaces are at $z=\pm H$, the vertically integrated <kinematic viscosity> satisfies
$$
\bar\nu\Sigma=\int_{-H}^H\rho\nu\,dz
=2H\frac\beta{1+\beta}\frac{c}{q^2\kappa_{\rm op}}.
$$
Use the previous result and the specified <Eddington accretion rate> convention,
$$
\dot M_E=\frac{L_E}{\eta c^2}
=\frac{4\pi GM}{\eta\kappa_{\rm op}c},\qquad \eta=\frac1{16},
$$
to obtain the full thickness
$$
\boxed{2H=\frac{64}{3}\frac{1+\beta}{\beta}\,q^2f\,
\frac{\dot M}{\dot M_E}\frac{GM}{c^2}}.
$$
The <constant-pressure-ratio vertical disk model> therefore has thickness proportional to the inward <accretion rate>. This is a full surface-to-surface thickness, rather than a density <disk scale height>. Its use as a <thin disk> requires $2H\ll r$.
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