= Solution
Use $N$ turns of mean radius $a\simeq1.05\,\mathrm{mm}$, including the wire radius, on a light nonmagnetic rod. With wire radius $r_w$, length $l$ and <density> $\rho$, neglecting helical pitch gives
$$
m_B=NI\pi a^2,\qquad M=2\pi aN\rho\pi r_w^2,
\qquad J=M\left(\frac{l^2}{12}+\frac{a^2}{2}\right).
$$
The last expression is the transverse central <moment of inertia> of the uniformly distributed turns. The <magnetic torque> is $-m_BB_h\sin\theta$, so the <small-angle magnetic dipole oscillation> obeys $J\ddot\theta+m_BB_h\theta=0$. Hence
$$
\boxed{T=2\pi\sqrt{\frac{2\pi\rho r_w^2}{IaB_h}
\left(\frac{l^2}{12}+\frac{a^2}{2}\right)}\simeq4.0\,\mathrm s.}
$$
The number of turns cancels. Taking the turn radius as $1.00\,\mathrm{mm}$ instead gives $4.04\,\mathrm s$, consistent at the accuracy of the thin-winding model. The rod's inertia, damping and lead-wire <torque> are neglected.
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