The correct answer is D.

The weight of observation i-days ago is given by:

$$w(i) = \frac{\lambda^{i-1}(1 - \lambda)}{1 - \lambda^n}$$

Where n is the number of days in the historical window and $\theta$ is the control rate of the memory decay,

Therefore:

$$W(13) = \frac{0.8^{13-1}(1 - 0.8)}{1 - 0.8^{300}} \approx 0.01374$$