$\begin{array}{l}\text{Here, }\sqrt{\text{19683}}_3\text{ = ?}\times \text{3}\\ \\ \Rightarrow \sqrt{\text{3}\times \text{3}\times \text{3}\times \text{3}\times \text{3}\times \text{3}\times \text{3}\times \text{3}\times \text{3}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} =? x 3}\\ \\ \text{[}\because \text{19683=3}\times \text{3}\times \text{3}\times \text{3}\times \text{3}\times \text{3}\times \text{3}\times \text{3}\times \text{3}]\\ \\ \Rightarrow {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(3}}^9{\text{)}}^{\times \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=?\times 3\\ \\ \Rightarrow 3^3=3\times ?\Rightarrow ?=\frac{27}{3}=9\end{array}$

Hence Option E is correct.