I have the question "Using the equation and information below, derive the units for the permittivity of free space , $\epsilon_{_0}$.
$V = \dfrac 1 {4 \pi \epsilon_{_0}} ~ \dfrac Q r$
$V$ is voltage in volts, $Q$ is charge in coulombs, $r$ is distance in metres, $4$ and $\pi$ are numbers and have no units."
From this information do I need to rearrange the formula to make e0 the subject and then workout the si units of each value?
When I checked the solutions the answer to this is $Q/Vm$ or $QV^{-1}m^{-1}$.
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".