= Solution
For slow gyroscopic <precession>, the gravitational <torque> turns the spin <angular momentum>: $Mg\ell\sin\theta=\Omega I_3\omega\sin\theta$, with $I_3=MR^2/2$. Thus
$$
\Omega\simeq\frac{2g\ell}{R^2\omega},\qquad
\boxed{T_p\simeq\frac{\pi R^2\omega}{g\ell}
=1.61\,\mathrm s,\qquad\omega=40\pi\,\mathrm{s^{-1}}.}
$$
The inclination cancels at this order. More precisely, <uniform precession of a heavy symmetric top> satisfies
$$
\frac{I_1}{M}\cos\theta\,\Omega^2
-\frac{I_3}{M}\omega\Omega+g\ell=0,
\qquad \frac{I_1}{M}=\frac{R^2}{4}+\ell^2.
$$
Its slow root gives $T_p\simeq1.57\,\mathrm s$. \b[About $1.6\,\mathrm s$] is the slow-precession answer; arbitrary initial conditions can also produce <nutation> rather than exact steady <precession>.
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