A coil of inductance 350 mH and resistance 25 Ω is joined to a battery of 15 V, with no other…

Step 1. The time constant. $\tau = \dfrac{L}{R} = \dfrac{0.350\ \text{H}}{25\ \Omega} = 0.014\ \text{s} = 14\ \text{ms}$.

Step 2. The starting rate. At $t = 0$ the current is zero, so $\dfrac{dI}{dt} = \dfrac{\mathcal{E}}{L} = \dfrac{15\ \text{V}}{0.350\ \text{H}} = 42.9\ \text{A/s}$.

Step 3. The final current. $\dfrac{\mathcal{E}}{R} = \dfrac{15\ \text{V}}{25\ \Omega} = 0.60\ \text{A}$.

Step 4. At $20$ ms. $\dfrac{t}{\tau} = \dfrac{20}{14} = 1.43$. So $I = 0.60\ \text{A} \times \left(1 - e^{-1.43}\right) = 0.60\ \text{A} \times 0.760 = 0.456\ \text{A}$.