I would use dimensional analysis to solve this problem.
The prefix μ, micro-, means 1E-6. = 0.000001. The μmol is a number of particles 1 million times smaller than a mole, so 6.02E17 particles rather than Avogadro's #. Of course we do not need to use this number in our problem, but it is good to keep in mind what a mole is due to its sometimes abstract nature.
7.0 mg nicotine (6.164 μmol/1 mg) = 7.0 x 6.164 = 43.1 μmol or 43 with 2 sig figs
The mg unit cancels in the above calculation as we multiply the quantity in parenthesis by the mass of nicotine given, 7.0 mg.
I hope this helps! Feel free to ask any follow up questions. :)
