A cylindrical buoy floats upright in a lake, with 0.40 m of it under the water…

(a) In the lake. $T = 2\pi\sqrt{\dfrac{0.40\ \text{m}}{9.8\ \text{m/s}^2}} = 1.27\ \text{s}$.

(b) In the sea. The buoyant force must still equal the weight, so $h\rho$ stays the same. We want $h$, so $h = \dfrac{0.40\ \text{m}}{1.03} = 0.388\ \text{m}$. Then $T = 2\pi\sqrt{\dfrac{0.388\ \text{m}}{9.8\ \text{m/s}^2}} = 1.25\ \text{s}$.