Answer
$t=\dfrac{L}{R} \ln (2)$ seconds
Work Step by Step
Consider $\dfrac{di}{dt}+(\dfrac{R}{T}) i=\dfrac{V}{L}$
Recall the formula: $i=(\dfrac{V}{R})(1-e^{-(\frac{R}{L})t})$
When the value of steady current becomes: $i=(\dfrac{1}{2})(\dfrac{V}{R})$, then we get $(\dfrac{1}{2})(\dfrac{V}{R})=(\dfrac{V}{R})(1-e^{-(\frac{R}{L})t})$
This implies that $\ln (1)-\ln (2)=\ln e^{-(\frac{R}{L})t}$
or, $-\ln (2)=\ln e^{-(\frac{R}{L})t}$
Hence, $t=\dfrac{L}{R} \ln (2)$ seconds
