= Solution
Choose a coordinate $z$ measured from the sphere centre towards increasing distance from the block, so an atom in the sphere is at height $L+z$. Summing the atom–block <potential energy> with the sphere's <number density> gives
$$
U(L)=-\frac{B\pi n^2}{6}\int_{|\boldsymbol x|<R}\frac{d^3x}{(L+z)^3}.
$$
For $R/L\ll1$, a uniformly valid <Taylor series> over the sphere is
$$
\frac1{(L+z)^3}=\frac1{L^3}\left[1-\frac{3z}{L}+\frac{6z^2}{L^2}-\frac{10z^3}{L^3}+O\!\left(\frac{z^4}{L^4}\right)\right].
$$
The odd moments vanish by reflection symmetry. The sphere volume is $V_s=4\pi R^3/3$, and its second moment is
$$
\int z^2d^3x=\pi\int_{-R}^{R}z^2(R^2-z^2)dz=\frac{4\pi R^5}{15}=\frac{V_sR^2}{5}.
$$
Consequently
$$
\boxed{U(L)=-\frac{2\pi^2Bn^2R^3}{9L^3}
\left[1+\frac65\frac{R^2}{L^2}+O\!\left(\frac{R^4}{L^4}\right)\right].}
$$
The first finite-radius correction is therefore $-4\pi^2Bn^2R^5/(15L^5)$. There is no relative $R/L$ correction because the first moment about the centre is zero. The leading term is just the number of sphere atoms times the potential at its centre; the correction increases the attraction because the nearer atoms contribute more strongly than the farther atoms lose.
This is the <far-field sphere-plane dispersion energy>, with <Hamaker constant> $A_H=\pi^2Bn^2$. It must be distinguished from the near-contact <sphere-plane dispersion force>, whose small parameter is the surface gap divided by the radius. Here $L$ is explicitly the centre-to-plane distance.
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