= Solution
In the small-body approximation, write $\chi_{\rm eff}=\chi/(1+N_d\chi)$, with $\chi=\mu-1$. The <induced magnetic dipole> moment is $\mathbf m=VM\widehat{\mathbf z}=V\chi_{\rm eff}H_0\widehat{\mathbf z}$. Its effective interaction <potential energy> at fixed source current is
$$
U_{\rm eff}(z)=-\frac12\mu_0V\chi_{\rm eff}H_0(z)^2.
$$
The factor $1/2$ accounts for inducing the moment: differentiating $-m(z)B_0(z)$ as if the moment were permanent would double the force. Equivalently, the instantaneous dipole force is $F_z=m\,dB_0/dz$. Thus the <force on a small linearly magnetizable body> is
$$
F_z=\mu_0V\chi_{\rm eff}H_0\frac{dH_0}{dz}.
$$
From part (b), $dH_0/dz=-3a^2Iz/[2(a^2+z^2)^{5/2}]$. Hence
$$
\boxed{\mathbf F=-\frac{3\mu_0V\chi_{\rm eff}a^4I^2z}{4(a^2+z^2)^4}\widehat{\mathbf z}.}
$$
For the ideal slender rod, $\chi_{\rm eff}\simeq\mu-1$. If the source's printed magnetisation is imposed instead, its corresponding conditional answer is
$$
\boxed{\mathbf F_{\rm printed}=-\frac{3\mu_0V(\mu-1)a^4I^2z}
{4\mu(a^2+z^2)^4}\widehat{\mathbf z}.}
$$
For $\mu>1$ and $z>0$, both expressions point toward the loop, where the applied <magnetic field> is stronger; for $z<0$ the direction reverses toward the loop plane. A material with <diamagnetism>, $\chi_{\rm eff}<0$, reverses this attraction. The force is in newtons, since a <magnetic dipole moment> times a field gradient has units $\mathrm{A\,m^2\,T\,m^{-1}}$. Both formulas assume <linear response>, negligible saturation and approximately uniform applied field across the sample.
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