Vector Products (Torque)  $$ T= |\underline r \space\times \underline F| $$ $$ OP =\underline r $$ $$ x = |\underline r|\cos\theta $$ $$ x = 0.8\cos40 $$ $$ x = 0.61\underline i $$ $$ y = |\underline r|\sin\theta $$ $$ y = 0.8\sin40 $$ $$ y = 0.51\underline j $$ $$ \underline r = 0.61\underline i + 0.51\underline j $$ $$ For\space \underline F $$ $$ x = |\underline F|\cos\theta $$ $$ x = 40\cos30 $$ $$ x = 20\sqrt{30}\underline i $$ $$ y = |\underline F|\sin\theta $$ $$ y = 40\sin30 $$ $$ y = 20\underline j $$ $$ \underline F = 20\sqrt{30}\underline i + 20\underline j $$ $$ T= |\underline r \space\times \underline F| $$ $$ T= |(0.61\underline i + 0.51\underline j)\times (20\sqrt{30}\underline i + 20\underline j)| $$ $$ T= |2.8\underline k| $$ $$ T= 2.8\space Nm $$ The answer is $37.6\space Nm\space $can someone please tell me how I have gone wrong or whether my whole method is wrong. Any help appreciated.

Another (probably a simpler) approach. Remember that

$$ |{\bf A}\times {\bf B}| = |{\bf A}| |{\bf B}| \sin \theta_{AB} $$

where $\theta_{AB}$ is the angle formed by ${\bf A}$ and ${\bf B}$. In your case

$$ |{\bf r}\times {\bf F}| = |{\bf r}| |{\bf F}| \sin \theta_{rF} = (0.8{\rm m})(50{\rm N})\sin(30^\circ+40^\circ) = 37.6~{\rm Nm} $$