Answer
Magnitude of $\vec F$ is $3.11\;kN$
and component of $\vec F$ along $v$ axis is $4.39\;kN$
Work Step by Step
Let $F_u$ and $F_v$ are tow components of $\vec F$ along the $u$ and $v$ axes.
The component along $u$ axis is given: $F_u=6\;kN$
Using the law of sines in the triangle formed by $F$, $F_u$, and $F_v$, we get
$\frac{F_u}{\sin105^\circ}=\frac{F_v}{\sin45^\circ}=\frac{F}{\sin30^\circ}$
$\therefore\;F=\frac{F_u}{\sin105^\circ}\times\sin30^\circ$
or, $F=\frac{6}{\sin105^\circ}\times\sin30^\circ\;kN\approx 3.11\;kN$
and $F_v=\frac{F_u}{\sin105^\circ}\times\sin45^\circ$
or, $F_v=\frac{6}{\sin105^\circ}\times\sin45^\circ\;lb\approx 4.39\;kN$
