= Solution
Let $a$ be the cube's side and $z$ a small upward displacement. Its displaced water <volume> decreases by $a^2z$, so the change in <buoyancy> is $-\rho_wga^2z$. With cube <mass> $m=0.9\rho_wa^3$,
$$
m\ddot z=-\rho_wga^2z,\qquad
\boxed{\omega^2=\frac{g}{0.9a},\qquad T=2\pi\sqrt{\frac{0.9a}{g}}=0.426\,\mathrm s.}
$$
The displacement must be small enough that the upper face remains above water and the bottom remains submerged. This <buoyancy> estimate neglects fluid <added mass>, <damping> and melting.
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