Answer
$1.1mol$
Work Step by Step
We can determine the required number of moles as follows:
$V=(2\times 10^{-2}m)^3=8\times 10^{-3}m^3$
and $M=\rho V$
$\implies M=(8920Kg/m^3)(8\times 10^{-6}m^3)=0.07136Kg=71.36g$
Now $n=\frac{M}{M_{mol}}$
We plug in the known values to obtain:
$n=\frac{1mol}{64g}(71.36g)$
$\implies n=1.1mol$
