$解：(1)G = mg = 5.4\ kg×10\ N/kg = 54\ N$
(2)$p=\frac{F}{S}=\frac{54\ N}{8×10^{-3}\ m^{2}} = 6.75×10^{3}\ Pa$
$(3)V = Sh = 8×10^{-3}\ m^{2}×0.25\ m = 2×10^{-3}\ m^{3}，$
$\rho=\frac{m}{V}=\frac{5.4\ kg}{2×10^{-3}\ m^{3}} = 2.7×10^{3}\ kg/m^{3}$
因为$2.7×10^{3}\ kg/m^{3}＞2.6×10^{3}\ kg/m^{3}，$所以合格
