Chapter 1 - Chemistry and Measurement - Questions and Problems - Page 39: 1.168

Please see the work below.

We know that $V=area \space thickness$ $V=5500000\times 6000\times (\frac{5280}{1})^2\times (\frac{12}{1})^2\times (\frac{2.54}{1})=2.605\times 10^{22}cm^3$ $mass=density\times volume$ $mass=0.917\times2.605=2.38\times 10^{22}g $