1. Discharge equations for a capacitor

When a capacitor discharges through a resistor $R$, the charge at time $t$ is

$Q(t)=Q_0 e^{-t/RC}$

where $Q_0$ is the initial charge.

The energy stored in the capacitor is

$U(t)=\frac{Q(t)^2}{2C}$

So,

$U(t)=U_0 e^{-2t/RC}$

where $U_0=\dfrac{Q_0^2}{2C}$ is the initial energy.

2. Find $t_1$ from energy becoming half

Given that in time $t_1$, energy becomes half of initial value:

$U(t_1)=\frac{U_0}{2}$

Using

$U(t)=U_0 e^{-2t/RC}$

we get

$U_0 e^{-2t_1/RC}=\frac{U_0}{2}$

$e^{-2t_1/RC}=\frac{1}{2}$

Taking natural log,

$-\frac{2t_1}{RC}=\ln\left(\frac{1}{2}\right)=-\ln 2$

$\frac{2t_1}{RC}=\ln 2$

$t_1=\frac{RC}{2}\ln 2$

3. Find $t_2$ from charge becoming one eighth

Given that in time $t_2$, charge becomes one eighth of initial value:

$Q(t_2)=\frac{Q_0}{8}$

Using

$Q(t)=Q_0 e^{-t/RC}$

we get

$Q_0 e^{-t_2/RC}=\frac{Q_0}{8}$

$e^{-t_2/RC}=\frac{1}{8}$

Taking natural log,

$-\frac{t_2}{RC}=\ln\left(\frac{1}{8}\right)=-\ln 8=-3\ln 2$

$t_2=3RC\ln 2$

4. Compute the ratio $t_1/t_2$

$\frac{t_1}{t_2}=\frac{\frac{RC}{2}\ln 2}{3RC\ln 2}$

Cancel $RC\ln 2$:

$\frac{t_1}{t_2}=\frac{1/2}{3}=\frac{1}{6}$

5. Check options

The correct option is

$\boxed{\frac{1}{6}}$

So, Option D is correct.