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To get the conversion factors:

$1000kg=1Mg$

$1{m}^{3}={10}^{6}c{m}^{3}$

$1c{m}^{3}=1mL$

$1mL={10}^{3}\mu L$

Now we can substitute with conversion factors to get

$\rho ={10}^{18}×\frac{kg}{{m}^{3}}×\frac{1×{m}^{3}}{{10}^{6}×c{m}^{3}}×\frac{1×c{m}^{3}}{1×mL}×\frac{1×mL}{{10}^{3}×\mu L}×\frac{1×Mg}{1000×kg}$

cancel similar units to het that

$\rho ={10}^{6}\frac{Mg}{\mu L}$