Answer
(a) $7.26\times 10^{-4}$ kg
(b) $93.7\%$
(c) 13.16 kg/m$^3$
Work Step by Step
(a) Dimensions of the brick in SI units:
$h=2 \mathrm{in.}=0.05$m
$b=4 \mathrm{in.}=0.10$m
$l=8 \mathrm{in.}=0.20$m
Dimensions of the cylindrical hole:
$r=1\mathrm{cm}=0.01$m
$H=8\mathrm{in.}=0.20$m
Density of lead, $\rho_{lead}=11.4 \times$ Density of water $= 11.4\times 1000=11400 \,\mathrm{kg/m}^3$
Mass of the lead removed from the brick is just the mass of the cylindrical hole: $m_{cylinder}=\rho_{lead}V_{cylinder}=11365.8\times π r^2 H=11400\times 3.14\times 0.01^2\times 0.20=0.7 \mathrm{kg}$
Hence mass of the lead removed from the brick is $0.7\mathrm{kg}$
(b) $\mathrm{Percentage \,of \,remaining \,lead}={\mathrm{Volume\, of\, hollowed\, brick}\over \mathrm{Volume \,of \,original \,brick}}\times 100$
$={\mathrm{Volume\, of\, original\, brick-Volume\, of\, cylindrical\, hole}\over \mathrm{Volume \,of \,original \,brick}}\times 100$
$={lbh-π r^2H\over lbh}\times 100$
$={0.05\times 0.10\times 0.20-3.14\times 0.01^2\times 0.20\over 0.05\times 0.10\times 0.20}\times 100$
$={0.001-0.00006\over 0.001}\times 100$
$={0.00094\over 0.001}\times 100$
$=0.94\times 100$
$=94\%$
About $94\%$ of the original lead remains in the brick.
(c) Mass of the hollowed brick, $m_b=\rho_{lead}\times \mathrm{Volume\, of\, hollowed\, brick}$
$=11400\times (lbh-π r^2 H)$
$=11400\times 0.00094$
$=11\mathrm{kg}$
Mass of plastic cylinder, $m_p=\rho_{plastic}\times \mathrm{Volume\, of\, cylindrical\, hole}$
$=2\times 1000\times 0.00006$
$=0.1\mathrm{kg}$
Total mass of the brick after fabrication, $M=m_b+m_p=11+0.1=11.1\mathrm{kg}$
Total volume of the brick after fabrication, $V= \mathrm{Total \,volume \,of \,original \,brick}=lbh=0.001\mathrm{m}^3$
Overall density of the brick-plastic combination$=\frac{M}{V}={11.1\over 0.001}=11100\mathrm{kg/m}^3$
