I believe the quoted passage contains an error.
The EC standard for lead in the atmosphere is 2 µg/m$^{−3}$. Express this value in scientific notation in g/cm$^{−3}$, to an appropriate number of significant figures.
It should either be 2 µg m$^{−3}$ or 2 µg/m$^3$, both of which have the same meaning, but not 2 µg/m$^{−3}$. Likewise, g/cm$^{−3}$ should either be g cm$^{−3}$ or g/cm$^3$. The negative exponent is just an alternative notation for division.
Think of µg/m$^3$ as "micrograms per cubic meter". This is a density. So a higher number means a greater density of lead. Visualize 2 µg of substance spread out over the volume of a cube with dimensions 1m $\times$ 1m $\times$ 1m.
Now as a warmup to doing unit conversions, ask yourself whether 2 µg/m$^3$ is a higher or lower density than 2 µg/cm$^3$. If you think about it, you will realize that the former represents 2 µg of substance spread out over a large cube with dimensions 1m $\times$ 1m $\times$ 1m, while the latter represents the same amount of substance spread out over a tiny cube with dimensions 1cm $\times$ 1cm $\times$ 1cm. So 2 µg/cm$^3$ is the higher density since the same amount of substance is crammed into a much smaller volume.
Now lets go back to the original density, 2 µg/m$^3$, and try to write it, first in units of µg/cm$^3$, and then in units of g/cm$^3$. A cube with dimensions 1m $\times$ 1m $\times$ 1m is the same as a cube with dimensions 100cm $\times$ 100cm $\times$ 100cm. Thinking about that cube as composed of many tiny 1cm $\times$ 1cm $\times$ 1cm cubes, how many of those tiny cubes are there? The answer is $100\times100\times100=100^3=10^6=1,000,000$. So if the big cube contains 2 µg of substance, which we imagine to be uniformly distributed, how much substance does each tiny cube contain? Since there are 1,000,000 tiny cubes, each contains $1/1000,000$ of the total amount of substance, or $2\times10^{-6}$ µg. Therefore, a density of 2 µg/m$^3$ is the same as a density of $2\times10^{-6}$ µg/cm$^3$. This insight can be encoded in the following algebraic manipulation:
$$
2\frac{{\mu}\text{g}}{\text{m}^3}=2\frac{{\mu}\text{g}}{\text{m}^3}\left(\frac{1\,\text{m}}{100\,\text{cm}}\right)^3=2\frac{{\mu}\text{g}}{\text{m}^3}\frac{\text{m}^3}{1,000,000\,\text{cm}^3}=\frac{2}{1,000,000}\frac{{\mu}\text{g}}{\text{cm}^3}.
$$
In the first step we are free to multiply by (1 m/100 cm) since that's just a way of writing 1.
Now let's deal with the conversion from µg to g. One µg is a tiny amount of material - just 1 millionth of a gram. If you have 1 µg of material in a certain volume, that's the same as having 1/1,000,000 g of material in the same volume. This insight allows us to complete the calculation:
$$
\frac{2}{1,000,000}\frac{{\mu}\text{g}}{\text{cm}^3}=\frac{2}{1,000,000}\frac{{\mu}\text{g}}{\text{cm}^3}\frac{\frac{1}{1,000,000}\text{g}}{\mu\text{g}}=\frac{2}{10^{12}}\frac{\text{g}}{\text{cm}^3}=2\times10^{-12}\frac{\text{g}}{\text{cm}^3}.
$$
Again we are allowed to multiply by (1/1,000,000 g)/(1 µg) in the first step since that's just equal to 1.
