= Solution
For a normal from medium 1 to 2, the magnetic <boundary conditions> are
$$
\boxed{\widehat{\mathbf n}\cdot(\mathbf B_2-\mathbf B_1)=0,\qquad
\widehat{\mathbf n}\times(\mathbf H_2-\mathbf H_1)=\mathbf K_{\rm free}.}
$$
Without a free surface current, tangential $H$ is continuous. To obtain the <magnetization of a permeable sphere> of <radius> $R$ and permeability $\mu_0(1+\chi)$, choose <magnetic scalar potentials>
$$
\varphi_{\rm in}=-H_i r\cos\theta,\qquad
\varphi_{\rm out}=-H_0r\cos\theta+C\cos\theta/r^2.
$$
Tangential continuity gives $H_i=H_0-C/R^3$, and normal $B$ continuity gives $(1+\chi)H_i=H_0+2C/R^3$. Hence
$$
\boxed{\mathbf H_i=\frac{\mathbf H_0}{1+\chi/3},\qquad
\mathbf M=\frac{\chi\mathbf H_0}{1+\chi/3}.}
$$
The fields are uniform inside. For negative susceptibility, <magnetization> is antiparallel to the applied field; the scalar formula retains its negative sign.
For a small weakly magnetic body, the induced <magnetic energy> is $U_m=-\chi VB^2/(2\mu_0)$. The factor one-half accounts for building the induced moment. The vertical <force> is $F_z=\chi VBB'/\mu_0$. Balancing <gravity> therefore gives
$$
\boxed{BB'=\frac{\mu_0\rho g}{\chi}.}
$$
A diamagnet needs a field decreasing in magnitude with height for an upward <force>. For <diamagnetic levitation in an inverse-height field>,
$$
BB'=-\frac{(B_0r_0)^2}{z^3},\qquad
\boxed{z_0=\left[\frac{-\chi(B_0r_0)^2}{\mu_0\rho g}\right]^{1/3}=0.0902\,\mathrm m\simeq9.0\,\mathrm{cm}.}
$$
The <force> falls as $z^{-3}$, giving a restoring vertical derivative at equilibrium. Full lateral stability would require the transverse field geometry, which is not specified.
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