To find leg $ b $ use formula:
$$ \tan \left( \beta \right) = \dfrac{ b }{ a } $$After substituting $ \beta = 30^o $ and $ a = 6 $ we have:
$$ \tan( 30^o ) = \dfrac{ b }{ 6 } $$ $$ \frac{\sqrt{ 3 }}{ 3 } = \dfrac{ b }{ 6 } $$$$ b = \frac{\sqrt{ 3 }}{ 3 } \cdot 6 $$$$ b = 2 \sqrt{ 3 } $$