Yes, we have:
$\displaystyle e^{-\Large\frac{R}{L}t}=\frac{1}{10}$
Or:
$\displaystyle e^{\Large\frac{R}{L}t}=10$
This implies (converting from exponential to logarithmic form):
$\displaystyle \frac{R}{L}t=\ln(10)$
$\displaystyle t=\frac{L}{R}\ln(10)$