Diagram: https://i.stack.imgur.com/d3jY5.jpg
The question is to find $F_2$
As you can see in the photo, I tried taking the horizontal components of $F_2$ and the force with magnitude $3$. I set $F_1$ as the sum of these forces then set $F_1=10$ because the system is in equilibrium and the other opposite force is $10$. This gave an answer that does not match any of the options.
On
It would be easier to consider vertical forces, that way you can ignore $F_1$, and if applicable, any error in such a calculation.
The total downward force is $3\sin(30)+4=5.5N$.
Therefore, the total vertical contribution from $F_2$ is $5.5N$.
Therefore, the magnitude of $F_2$ is $5.5\sqrt{2}$ N, or option B.
Hint:
In the vertical direction, we have $\dfrac{ F_2} {\sqrt {2}}-4-\dfrac32=0.$