= Solution
The printed formula has a physical problem for the specified geometry. Let $\chi=\mu-1$ be the linear <magnetic susceptibility>, where $\mu$ is the source's relative <magnetic permeability>. The <constitutive equation> is $M=\chi H_{\rm in}$, not in general $\chi H_0/\mu$.
For a small <uniformly magnetized ellipsoid> with its principal axis along the applied field, include its <demagnetizing field> through the SI <demagnetizing factor> $N_d$:
$$
H_{\rm in}=H_0-N_dM,\qquad M=\chi H_{\rm in}.
$$
Solving gives
$$
\boxed{M=\frac{\mu-1}{1+N_d(\mu-1)}H_0
=\frac{(\mu-1)a^2I}{2[1+N_d(\mu-1)](a^2+z^2)^{3/2}}.}
$$
A slender rod aligned along the field has $N_d\to0$ as its aspect ratio grows. Its ideal slender-rod limit at fixed $\mu$ is therefore
$$
\boxed{M_{\rm rod}\simeq(\mu-1)H_0
=\frac{(\mu-1)a^2I}{2(a^2+z^2)^{3/2}},}
$$
provided $N_d|\mu-1|\ll1$. A finite cylinder generally has nonuniform magnetization, so an ellipsoidal shape factor is an approximation, and its aspect ratio is needed for a more precise correction. Small sample length compared with the field-variation scale makes $H_0$ approximately uniform across it; that condition does not set the <demagnetizing factor> to one.
The source's displayed $M=(\mu-1)H_0/\mu$ instead follows from $N_d=1$, the thin-plate limit with field normal to the plate. It can also be obtained by imposing $B_{\rm in}=B_0$, but that identification is not justified for a thin rod parallel to the field. Consequently the requested extra $1/\mu$ cannot honestly be derived from the stated rod geometry. The next part gives both the shape-consistent result and the force conditional on using the printed magnetisation.
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