= Solution
For a statistically homogeneous and isotropic density contrast,
$$
\boxed{
\xi(r)=\langle\delta(\mathbf x)
\delta(\mathbf x+\mathbf r)\rangle,
\qquad r=|\mathbf r|
}.
$$
Using the <Fourier transform> convention
$$
\delta(\mathbf x)=\int\frac{d^3k}{(2\pi)^3}
\delta_{\mathbf k}e^{i\mathbf k\cdot\mathbf x}
$$
and translational invariance gives
$$
\langle\delta_{\mathbf k}\delta_{\mathbf k'}^*\rangle
=(2\pi)^3\delta_D(\mathbf k-\mathbf k')P(k),
$$
with
$$
\boxed{
P(k)=\int d^3r\,\xi(r)e^{-i\mathbf k\cdot\mathbf r}
}.
$$
At one point,
$$
\boxed{
\sigma^2=\xi(0)
=\frac1{2\pi^2}\int_0^\infty k^2P(k)\,dk
}.
$$
For a spherical top-hat volume $V=4\pi R^3/3$,
$$
\boxed{
\sigma^2(R)
=\frac1{V^2}\int_Vd^3x_1\int_Vd^3x_2\,
\xi(|\mathbf x_1-\mathbf x_2|)
}.
$$
The overlap of two radius-$R$ balls separated by $r\leq2R$ is
$$
V_{\rm ov}(r)=V\left(
1-\frac{3r}{4R}+\frac{r^3}{16R^3}
\right).
$$
Therefore
$$
\boxed{
\sigma^2(R)
=\frac3{R^3}\int_0^{2R}
r^2\xi(r)
\left(1-\frac{3r}{4R}
+\frac{r^3}{16R^3}\right)dr
}.
$$
For $\xi(r)=(r/r_0)^{-1.8}$,
$$
\sigma^2(R)=J_2\left(\frac{r_0}{R}\right)^{1.8},
\qquad
J_2=\frac{72}
{2^{1.8}(3-1.8)(4-1.8)(6-1.8)}
\simeq1.865.
$$
Thus
$$
\boxed{
\sigma_8
=\sqrt{J_2}\left(\frac58\right)^{0.9}
\simeq0.895
}.
$$
This calculation assumes that the observed galaxies trace the matter field with unit, scale-independent <galaxy bias>[galaxy bias], and that the fitted <power law> remains valid throughout the <spherical top-hat window function>[top-hat] separations.
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