Answer
3.7 time constants.
Work Step by Step
The current through the LR circuit after the battery is removed is given by the following equation, discussed in section 21-12.
$$I=I_{max}e^{-t/\tau}$$
$$\frac{I}{ I_{max}}= e^{-t/\tau} $$
Set I equal to 2.5 percent of its original value, and solve for the time t.
$$0.025= e^{-t/\tau}$$
$$t=-\tau ln(0.025)\approx 3.7\tau $$
It takes 3.7 time constants.