Answer
$4.36\;{\text{units of force}},\; 263.4^\circ$
Work Step by Step
To keep the knot stationary, the net force exerted on it must be zero.
So, the force applied by you must equal and be opposite the net force of the two other forces.
$$\vec F_{you}=-\vec F_{net}=-(3\cos 0^\circ +5\cos120^\circ)\hat i-(3\sin 0^\circ +5\sin120^\circ)\hat j$$
$$\vec F_{you}= - 0.5\;\hat i-4.33\;\hat j$$
The magnitude of $\vec F_{you}$ is given by applying the Pythagorean theorem.
$$|\vec F_{you}|=\sqrt{F_x^2+F_y^2}=\sqrt{(-0.5)^2+(-4.33)^2}=\color{red}{\bf 4.36}\;\rm units \;of\; force$$
and its direction is given by
$$\tan\alpha_D=\dfrac{D_y}{D_x}$$
Thus,
$$\alpha_F=\tan^{-1}\left[\dfrac{F_y}{F_x}\right]=\tan^{-1}\left[\dfrac{-4.33}{- 0.5}\right]=\bf 83.4^\circ$$
And since it is in the third quadrant,
$$\alpha_F=180^\circ +83.4^\circ =\color{red}{\bf 263.4^\circ}$$
