= Solution
Interpret the longitudinal <Dirac delta function> as the <Markov approximation for a random medium>. The two arguments on the left of the printed covariance must be $x_1$ and $x_2$. The transverse <covariance kernel> is $B(|\mathbf z_1-\mathbf z_2|)$. Here <statistical isotropy> means transverse isotropy: a medium with a distinguished longitudinal <white noise> direction and a smooth transverse covariance is not literally isotropic in all three directions. Also, ideal <Gaussian white noise> replaces the original finite-<variance> field; it cannot simultaneously satisfy a pointwise normalization $\mathbb E W^2=1$.
Write $Q(r)=\mu^2B(r)$ for the covariance strength of the <refractive index> fluctuations. Define the transverse <power spectrum> using the unnormalized forward <Fourier transform> printed in equation 3:
$$
S_F(\boldsymbol\nu)=\int_{\mathbb R^2}Q(|\mathbf r|)e^{-i\boldsymbol\nu\cdot\mathbf r}\,d\mathbf r,\qquad
Q(\mathbf r)=\frac1{(2\pi)^2}\int_{\mathbb R^2}S_F(\boldsymbol\nu)e^{i\boldsymbol\nu\cdot\mathbf r}\,d\boldsymbol\nu.
$$
This is the spectrum of the fluctuations, excluding the deterministic mean index. Equivalently it is the zero-longitudinal-frequency slice of the three-dimensional fluctuation spectrum before the <Markov approximation>. The angular integral is $2\pi J_0(\nu r)$, where $J_0$ is the <Bessel function of the first kind>. Consequently <Fourier inversion> and the <Hankel transform> give
$$
S_F(\nu)=2\pi\int_0^\infty Q(r)J_0(\nu r)r\,dr,\qquad
Q(r)=\frac1{2\pi}\int_0^\infty S_F(\nu)J_0(\nu r)\nu\,d\nu.
$$
The printed equations 4 and 5 omit these reciprocal factors, and the left-hand side of equation 5 should depend on $r$. They are instead a consistent order-zero <Hankel transform> pair if their $F$ denotes $S_H=S_F/(2\pi)$.
Using $J_0(0)=1$, the mean-field solution in the <Fourier transform> convention is
$$
\boxed{m(x,\mathbf z)=\exp\left[-\frac{k_0^2x}{4\pi}\int_0^\infty S_F(\nu)\nu\,d\nu\right](U_xE_0)(\mathbf z).}
$$
Here $U_x$ is the <Fresnel propagator>. Finiteness of the integral ensures a finite screen <wave phase> <variance>. If the quoted <power spectrum> uses the self-reciprocal <Hankel transform> convention, the same result reads
$$
\boxed{m(x,\mathbf z)=\exp\left[-\frac{k_0^2x}{2}\int_0^\infty S_H(\nu)\nu\,d\nu\right](U_xE_0)(\mathbf z).}
$$
These formulas describe identical media when $S_F=2\pi S_H$; assigning the same numerical function to both spectral conventions describes different covariance strengths.
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