A capacitor is discharging through a resistor R . Consider in time...

We know that in the discharging circuit the charge is given by,

$q = {q}_{0}e\dfrac{- t}{RC}$ $⇒ \dfrac{t}{RC} = \ln \left(\dfrac{{q}_{0}}{q}\right)$

In time ${t}_{1}$ the energy stored reduces to half. Hence, $\dfrac{{{q}_{1}}^{2}}{2C} = \dfrac{1}{2} \times \dfrac{{{q}_{0}}^{2}}{2C} ⇒ \dfrac{{q}_{0}}{{q}_{1}} = \sqrt{2}$. Therefore,

$\dfrac{{t}_{1}}{RC} = \ln \left(\sqrt{2}\right) = \dfrac{1}{2}\ln 2$.

In time ${t}_{2}$ the charge stored reduces to $\dfrac{1}{8}th$ of initial value. Hence, $\dfrac{{q}_{2}}{{q}_{0}} = \dfrac{1}{8}$. Therefore,

$\dfrac{{t}_{2}}{RC} = \ln \left(8\right) = 3\ln 2$.

Hence,

$\dfrac{{t}_{1}}{{t}_{2}} = \dfrac{\dfrac{1}{2}\ln 2}{3\ln 2} = \dfrac{1}{6}$