The given equation for the amplitude of the charge oscillating in a circuit is:

$$Q = Q_0 e^{- \frac{R t}{2 L} }$$

We need to find the time, $t$, at which charge amplitude decreases to $0.50 Q_0$. So, we set $Q = 0.50 Q_0$:

$$0.50 Q_0 = Q_0 e^{- \frac{R t}{2 L} }$$

Divide both sides by $Q_0$:

$$0.50 = e^{- \frac{R t}{2 L} }$$

Take the natural logarithm on both sides to solve for $t$:

$$\ln(0.50) = - \frac{R t}{2 L}$$

Recall that $\ln(0.50) = - \ln(2)$. Substituting the given value $\ln(2) = 0.693$, we get:

$$-0.693 = - \frac{R t}{2 L}$$

Remove the negative signs from both sides:

$$0.693 = \frac{R t}{2 L}$$

Now, solve for $t$:

$$t = \frac{2 L \cdot 0.693}{R}$$

Substituting the given values $R = 1.5 \Omega$ and $$L = 12 \mathrm{~mH} = 12 \times 10^{-3} \mathrm{~H}$$, we have:

$$t = \frac{2 \times 12 \times 10^{-3} \cdot 0.693}{1.5}$$

Calculate the numerator and denominator:

$$t = \frac{16.632 \times 10^{-3}}{1.5}$$

Finally, compute the value of $t$:

$$t = 11.09 \times 10^{-3} \mathrm{~s} = 11.09 \mathrm{~ms}$$

Therefore, the time at which the charge amplitude decreases to $0.50 Q_0$ is nearly 11.09 ms. Thus, the correct answer is:

Option B: 11.09 ms