Amount of DCl = 4 milimole

Amount of NaOD = 8 milimole

After reaction of acid and base 

Amount of NaOD remaining = 4 milimole

Total volume = 100 ml = 0.1 lit

$$\begin{gathered} so \left[OD\right] =4*{10}^{-3}/0.1=4*{10}^{-2} \left(1\right) \\ Given pD = 13.6 \\ so \log \left[D^{+}\right] =-13.6= \overline{14}.4 \\ \Rightarrow \left[D^{+}\right] =2.5*{10}^{-14} \left(2\right) \\ Now solve \end{gathered}$$