Conversion of grams to milliliters using kg/m3 density
Concerning the units in your equation none of them is correct (As all your formulations would end up in a square volume unit).

Definition of volumetric mass density

The density $ \rho $ is given as

$$ \rho = \frac{m}{V} $$

resulting in a mass and density dependent Volume as

$$ V = \frac{m}{\rho} $$



Direct way

Although you could calculate this by just substituting the physical quantities by the given values and units, and then take care of the unit conversion afterward:

$$ V = \frac{m}{\rho} = \frac{100\,\mathrm{g}}{681.9\,\mathrm{kg}\,\mathrm{m^{-3}}} =\\= \frac{100\,\mathrm{g}\,\mathrm{m^{3}}}{681.9\,\mathrm{kg}} = \frac{100\,\mathrm{g} \cdot \left(10^2\,\mathrm{cm}\right)^{3}}{681.9 \cdot (10^3\,\mathrm{g})} =\\= \frac{100 \cdot 10^6\,\mathrm{cm}^{3}}{681.9 \cdot 10^3} = \frac{100 \cdot 10^3}{681.9}\,\mathrm{cm}^{3} =\\\approx 146.65\,\mathrm{cm}^{3} \equiv 146.65\,\mathrm{ml} $$

I would suggest to convert the density from $ \mathrm{kg}\,\mathrm{m^{-3}}  $ to $ \mathrm{g}\,\mathrm{cm^{-3}} $ as a first step:



Easier way

Of course easiness is something subjective, but I believe this is easier to use.

Calculate conversion-factor for density

First the combined conversion-factor for the units-fraction is calculated

$$ \frac{\mathrm{kg}}{\mathrm{m}^3} \stackrel{\left[1\right]}{=} \frac{\left(10^3\,\mathrm{g}\right)}{\left(10^2\,\mathrm{cm}\right)^3} \stackrel{\left[2\right]}{=} \frac{10^3\,\mathrm{g}}{10^6\,\mathrm{cm}^3} \stackrel{\left[3\right]}{=} 10^{-3}\,\frac{\mathrm{g}}{\mathrm{cm}^3} $$


Start with substituting each units SI prefix:
Care has to be taken with the conversion of $ \mathrm{m}^3 $, since the power applies to the whole substituted expression ($ \mathrm{m}^3 = \left(10^2\,\mathrm{cm}\right)^3 $).
In the last step the exponential expressions are summarized ($ \frac{10^3}{10^6} = 10^{3-6} = 10^{-3} $).


Then calculate the new numeric value of the density in the target unit:

$$ \rho = 681.9\,\mathrm{kg}\,\mathrm{m^{-3}} \stackrel{\left[4\right]}{\equiv} 681.9 \cdot \left(10^{-3}\,\mathrm{g}\,\mathrm{cm}^{-3}\right) \stackrel{\left[5\right]}{=} 0.6819\,\mathrm{g}\,\mathrm{cm}^{-3} $$


Substitute the unit with the previously calculated expression
Multiply the numeric terms


Calculate the volume

Now the unit-adjusted density is used:

$$ V = \frac{m}{\rho} = \frac{100\,\mathrm{g}}{0.6819\,\mathrm{g}\,\mathrm{cm^{-3}}} = \frac{100}{0.6819}\,\mathrm{cm^{3}} \approx 146.65\,\mathrm{cm}^{3} \equiv 146.65\,\mathrm{ml}  $$



Conversion-factors

According to SI Prefixes the following applies:

$ 1\,\mathrm{m} \equiv 10^2\,\mathrm{cm} \\
1\,\mathrm{kg} \equiv 10^3\,\mathrm{g} $

And of course:

$ 1\,\mathrm{cm}^3 \equiv 1\,\mathrm{ml} $