= Solution
Let the <cylinder> length be $\ell=0.020\,\mathrm m$, cross-sectional <area> $A$, and the fluid <mass density> $\rho_f$. At <floating equilibrium>, <Archimedes' principle> gives
$$
mg=\rho_f gA\frac\ell2,\qquad m=\rho_f A\frac\ell2.
$$
For a small downward displacement $z$ from this <equilibrium>, the displaced fluid <volume> increases by $Az$. The additional upward <buoyant force> is $\rho_fgAz$, and <Newton's second law> becomes
$$
m\ddot z=-\rho_fgAz,\qquad\ddot z+\frac{2g}{\ell}z=0.
$$
This is <simple harmonic motion>, with <angular frequency> $\omega=\sqrt{2g/\ell}$ and <oscillation period>
$$
\boxed{T=2\pi\sqrt{\frac\ell{2g}}=2\pi\sqrt{\frac{0.020\,\mathrm m}{2(9.81\,\mathrm{m\,s^{-2}})}}=0.2006\,\mathrm s\simeq0.20\,\mathrm s.}
$$
The result for <vertical oscillations of a floating cylinder> is independent of its cross-sectional <area> and fluid <mass density> because those quantities cancel using the known immersion fraction. It neglects <drag>, surface effects and the extra fluid inertia from the motion; the <cylinder> must remain partly immersed throughout the small motion.
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