Answer
$\tau_c > \tau_a> \tau_b$
Work Step by Step
The time constant of an $LR$ circuit is $\quad\tau = \displaystyle \frac{L}{R}$
The time constant of each circuit is
$\tau_a = \displaystyle \frac{L}{R_{a}} = \frac{L}{R}$
$\tau_b = \displaystyle \frac{L}{R_{b}} = \frac{L}{R+R} = \frac{L}{2R}$
$\tau_c = \displaystyle \frac{L}{R_{c}} = \frac{L}{(\frac{1}{R}+\frac{1}{R})^{-1}} = \frac{L}{R/2} = 2\frac{L}{R}$
The magnitudes of the time constants, ranked from largest to smallest, are:
$\tau_c > \tau_a> \tau_b$
