Answer
$\frac{\Delta \Phi_B}{\Delta t}$ can be measured in volts.
Work Step by Step
We can verify the units of $\frac{\Delta \Phi_B}{\Delta t}$:
$\frac{Wb}{s} = \frac{kg~m^2~s^{-2}~A^{-1}}{s}$
$\frac{Wb}{s} = \frac{J}{s~A}$
$\frac{Wb}{s} = \frac{J}{s~(C/s)}$
$\frac{Wb}{s} = \frac{J}{C}$
$\frac{Wb}{s} = V$
Therefore, $\frac{\Delta \Phi_B}{\Delta t}$ can be measured in volts.
