= Solution
Let the stationary <scattering potential> fluctuation have <autocorrelation function of a random field>
$$
C_W(\boldsymbol\rho)=\langle W(\mathbf r')W(\mathbf r'+\boldsymbol\rho)\rangle.
$$
Because $W$ is centered, this is also its <covariance function>. Multiplying the two real <wave phase> integrals and taking expectations gives
$$
\boxed{\langle\varphi(\mathbf r)^2\rangle=\iint b(\mathbf r,\mathbf r')b(\mathbf r,\mathbf r'')C_W(\mathbf r''-\mathbf r')\,d\mathbf r'\,d\mathbf r''.}
$$
The two minus signs cancel. This is the <wave phase> <variance>, since the <wave phase> mean is zero. More generally the <phase covariance in the first Rytov approximation> replaces the first kernel by $b(\mathbf r_1,\mathbf r')$ and the second by $b(\mathbf r_2,\mathbf r'')$. Finite observation/scattering windows, or suitable weighted-integrability hypotheses, make these double integrals well-defined in a stationary infinite-medium model.
The correlation needed here is that of the <scattering potential> fluctuation. With the printed $V=n^2-1$, the <covariance of a squared random field> is
$$
C_W(\boldsymbol\rho)=\langle n(\mathbf r')^2n(\mathbf r'+\boldsymbol\rho)^2\rangle-\langle n^2\rangle^2.
$$
It involves a fourth moment of $n$, so its value is not generally determined by the ordinary two-point correlation $C_n=\langle n(\mathbf r')n(\mathbf r'+\boldsymbol\rho)\rangle$ alone. If one additionally assumes a zero-mean <Gaussian random field>, <Isserlis theorem> yields $C_W=2C_n^2$. That assumption is not printed and must not be inserted silently. Alternatively, for a physical weak fluctuation $n_{\rm phys}=1+\eta$, $W\simeq2\eta$ gives $C_W\simeq4C_\eta$. \b[The general answer uses $C_W$; either reduction to a <refractive index> two-point correlation requires an extra assumption.]
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