解：$(1)$已知铁的密度$ρ_{铁}=7.9×10^3\ \mathrm {kg/m}^3=7.9\ \mathrm {g/cm}^3，$由$ρ=\frac {m}{V}$得铁球中铁的体积$V_{铁}=\frac {m}{ρ_{铁}}=\frac {158\ \mathrm {g}}{7.9\ \mathrm {g/cm}^3}=20\ \mathrm {cm}^3；$铁球内注入水的质量$m_{水}=m_{总}−m=308\ \mathrm {g}−158\ \mathrm {g}=150\ \mathrm {g}，$由$ρ=\frac {m}{V}$得空心部分的体积$V_{空心}=V_{水}=\frac {m_{水}}{ρ_{水}}=\frac {150\ \mathrm {g}}{1\ \mathrm {g/cm}^3}=150\ \mathrm {cm}^3=1.5×10^{-4}\ \mathrm {m^3}；$空心铁球的总体积$V=V_{铁}+V_{空心}=20\ \mathrm {cm}^3+150\ \mathrm {cm}^3=170\ \mathrm {cm}^3$
$(2)$装满液体后液体的体积$V_{液}=V_{空}=1.5×10^{-4}\ \mathrm {m^3}，$由$ρ=\frac {m}{V}$得装入液体的质量$m_{液}=ρ_{液}V_{液}=0.8×10^3\ \mathrm {kg/m}^3×1.5×10^{-4}\ \mathrm {m^3}=0.12\ \mathrm {kg}=120\ \mathrm {g}$
$(3)$设小石块的密度是$ρ_{石}，$体积是$V_{石}，$注入的水的体积是$V_{水}'，$则有$m_{总}=m_{球}+m_{石}+m_{水}'，$代入数据有$458\ \mathrm {g}=158\ \mathrm {g}+250\ \mathrm {g}+m_{水}'；$解得$m_{水}'=50\ \mathrm {g}，$由$ρ=\frac {m}{V}$可得，水的体积$V_{水}'=\frac {m_{水}'}{ρ_{水}}=\frac {50\ \mathrm {g}}{1\ \mathrm {g/cm}^3}=50\ \mathrm {cm}^3，$已经求得空心部分的体积为$150\ \mathrm {cm}^3，$则小石块的体积$V_{石}=V_{空}−V_{水}'=150\ \mathrm {cm}^3−50\ \mathrm {cm}^3=100\ \mathrm {cm}^3；$小石块的密度$ρ_{石}=\frac {m_{石}}{V_{石}}=\frac {250\ \mathrm {g}}{100\ \mathrm {cm}^3}=2.5\ \mathrm {g/cm}^3$
