$\log$2 \space \log{\sqrt{2}} \log_3 81 =

1. 1

2. 2

3. $\dfrac{1}{\sqrt{2}}$

4. $\dfrac{1}{2}$

Given,

$\Rightarrow \log$2 \space \log{\sqrt{2}} \space \log3 81 \\[1em] \Rightarrow \log2 \space \log{\sqrt{2}} \space \log3 3^4 \\[1em] \Rightarrow \log2 \space \log{\sqrt{2}} \space 4\log3 3 \\[1em] \Rightarrow \log2 \space (\log{\sqrt{2}} \space 4) \\[1em] \Rightarrow \log2 \space \Big(\dfrac{\log \space 4}{\log \space \sqrt{2}}\Big) \\[1em] \Rightarrow \log2 \space \Big(\dfrac{\log \space 2^2}{\log \space \sqrt{2}}\Big) \\[1em] \Rightarrow \log2 \space \Big(\dfrac{2\log \space 2}{\dfrac{1}{2}\log \space 2}\Big) \\[1em] \Rightarrow \log2 \space {\Big(\dfrac{2}{\dfrac{1}{2}}\Big)} \\[1em] \Rightarrow \log2 \space 4 \\[1em] \Rightarrow \log2 \space 2^2 \\[1em] \Rightarrow 2\log2 \space 2 \\[1em] \Rightarrow 2.

Hence, option 2 is the correct option.