Length of the cube, $L =1.2 \times 10^{-2} m$
$\therefore $ Volume of the cube,
$V = L ^{3}=\left(1.2 \times 10^{-2}\right) \times\left(1.2 \times 10^{-2}\right) \times\left(1.2 \times 10^{-2}\right) m ^{3}$
$\Rightarrow V =1.728 \times 10^{-6} m ^{3}$
But as the length contains only $2$ significant digits,
thus volume must also contain $2$ significant digits.
Hence, volume of the cube is $1.7 \times 10^{-6} m ^{3}$