Mad 9

$[{ x }^{ 3 }=({ { 10 }^{ 2016 }-1 })^{ 3 }\Rightarrow { 10 }^{ 6048 }-3\times { 10 }^{ 4032 }+{ 10 }^{ 2016 }-1=999...99700000...2999...99\\ 999...997......\quad have\quad 2016-1\quad 2015\quad 9's\quad \quad and\quad .....29999...999\quad have\quad 2016\quad 9's\\ \therefore 2015+2016=4031]$

$9^3=729$

$99^3=970299$

$999^3=997002999$

$9999^3=999700029999$

By inspection, if the number of 9s in the original number (consisting of only 9s) is $n$ , then the cube of that number will have $2n-1$ 9s. In this case, the number of 9s in the cube is $2\times2016-1=\boxed{4031}$