In a L–R decay circuit, the initial current at t = 0 is I. Find the total charge that has flown through the resistor till the energy in the inductor has reduced to one–fourth its initial value.

Energy in inductor =(1/2)LI$^2$

So final current would be I/2

Using kirchoff equation

$$L\frac{di}{dt} + ir =0$$

Interating it , I got

$$ln(\frac{I}{I_0 }=-\frac{R}{L}T$$

Where I$_0$= I/2

ln(2)=(-R/L)T

How this can be possible for this equation to satisfy T must be negative , that is impossible .

So you get $ln(\frac{I}{I_0})=ln(\frac12)=-ln(2)=-\frac{R}{L}T$.