Yes, solving for $\epsilon_0$ first is a good idea:
$$V = \frac{1}{4\pi\epsilon_0}\frac{Q}{r} \\\ \epsilon_0 = \frac{1}{4\pi V}\frac{Q}{r} \\\ [\epsilon_0] = \frac{1}{\textrm{V}}\frac{\textrm{C}}{\textrm{m}} = \textrm{C} \textrm{V}^{-1}\textrm{m}^{-1}$$
Read out, $\epsilon_0$ has units of "Coulombs per volt per meter".