Answer
$4.8\times 10^{23}atoms$
Work Step by Step
The required number of atoms can be determined as follows:
$V=(2.0\times 10^{-2}m)^3=8.0\times 10^{-6}m^3$
$M=\rho V$
$\implies M=(2700Kg/m^3)(8\times 10^{-6}m^3)=0.0216Kg=21.6g$
Now $N=N_A\cdot \frac{1}{M_{mol}}\cdot M$
We plug in the known values to obtain:
$N=(\frac{6.02\times 10^{23}atoms}{1mol})(\frac{1mol}{27g})(21.6g)$
This simplifies to:
$N=4.8\times 10^{23}atoms$
