a b c d

c

$ I = I_0 e^{\frac{-t}{\tau}}$ $\Rightarrow \frac{dI}{dt} = -\frac{I_0}{\tau} e^{\frac{-t}{\tau}}$ $ \text{At } t = 0 , \frac{dI}{dt} = \frac{I_0}{\tau} =  \frac{I_0 R}{L} =  \frac{V}{L} = 2.5 Amp/s$