$\frac{3}{{\sqrt 2  + \sqrt 3 }} = \frac{3}{{(\sqrt 2  + \sqrt 3 )}} \times \frac{{(\sqrt 2  - \sqrt 3 )}}{{(\sqrt 2  - \sqrt 3 )}} = \frac{{3(\sqrt 2  - \sqrt 3 )}}{{ - 1}}$

$\frac{3}{{\sqrt 3  + \sqrt 4 }} = \frac{3}{{(\sqrt 3  + \sqrt 4 )}} \times \frac{{(\sqrt 3  - \sqrt 4 )}}{{(\sqrt 3  - \sqrt 4 )}} = \frac{{3(\sqrt 3  - \sqrt 4 )}}{{ - 1}}$

$\frac{3}{{\sqrt {31}  + \sqrt {32} }} = \frac{3}{{(\sqrt {31}  + \sqrt {32} )}} \times \frac{{(\sqrt {31}  - \sqrt {32} )}}{{(\sqrt {31}  - \sqrt {32} )}} = \frac{{3(\sqrt {31}  - \sqrt {32} )}}{{ - 1}}$

$\begin{gathered}
  B = \frac{{3(\sqrt 2  - \sqrt 3 )}}{{ - 1}} + \frac{{3\sqrt 3  - \sqrt 4 }}{{ - 1}} + ... + \frac{{3(\sqrt {31}  - \sqrt {32} )}}{{ - 1}} \Rightarrow  \hfill \\
   \Rightarrow B = 3\left( {( - \sqrt 2  - \sqrt 3 ) + ( - \sqrt 3  + \sqrt 4 ) + ...( - \sqrt {31}  + \sqrt {32} )} \right) \hfill \\ 
\end{gathered} $

$\begin{gathered}
   \Rightarrow B = 3( - \sqrt 2  + \sqrt 2  - \sqrt 3  + \sqrt 4  + ... - \sqrt {31}  + \sqrt {32} ) \hfill \\
   \Rightarrow B = 3( - \sqrt 2  + \underbrace {\sqrt {32} }_{\sqrt {16 \times 2} }) = 3( - \sqrt 2  + 4\sqrt 2 ) \hfill \\ 
\end{gathered} $

$ \Rightarrow B = 3(3\sqrt 2 ) = 9\sqrt 2  \Rightarrow {B^2} = {(9\sqrt 2 )^2} = 162$