$ 解：(1)F_浮=G=mg=0.21\ \text {kg}×10\ \text {N/kg}=2.1\ \text {N}$
$ (2)G_总=F_浮'=ρ_水gV_排=1.0×10^{3}\ \text {kg/m}^3×10\ \text {N/kg}×3×10^{-3}×0.11\ \text {m}^3=3.3\ \text {N} $
$ G_A=G_总-G=3.3\ \text {N}-2.1\ \text {N}=1.2\ \text {N}$
$ m_A=\frac {G_A}{g}=\frac { 1.2\ \text {N}}{10\ \text {N/kg}}=0.12\ \text {kg}=120\ \text {g}$
$ (3)ρ_A=\frac {m_A }{V}=\frac {0.12\ \text {kg}}{ 1.5×10^{-5}\ \text {m}^3}=8×10^{3}\ \text {kg/m}^3 $
