Explanation

(c) Given, self inductance, $L = 1.8 \times 10^{- 4}H$
Resistance, $R = 6\Omega$
When self inductance and resistance is broken up into identical coils.
Then, self inductance of each oil
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ \frac{1.8 \times 10^{- 4}}{2}H$
Resistance of each oil
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ \frac{6\Omega}{2} = 3\Omega$
Coil are then connected in parallel
$\therefore\ \ \ \ \ \ \ \ L^{'} = \frac{\frac{1.8}{2} \times 10^{- 4} \times \frac{1.8}{2} \times 10^{- 4}}{\frac{1.8}{2} \times 10^{- 4} + \frac{1.8}{2} \times 10^{- 4}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 0.45 \times 10^{- 4}H$
and $\ \ R^{'} = \frac{3 \times 3}{3 + 3} = 1.5\Omega$
Time constant$\ = \frac{L^{'}}{R'}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ \frac{0.45 \times 10^{- 4}}{1.5} = 0.3 \times 10^{- 4}s$