Answer
$2.62\;\rm m$
Work Step by Step
Since the knife cuts the rope when the ball is just at the bottom of its circular path, we need to find its speed at this point when the tension is, as given, 5 N.
So we need to draw a force diagram of the ball at this point and then apply Newton's second law, as seen below.
$$\sum F_r=T-mg=ma_r=m\dfrac{v^2}{r}$$
where $r$ here is the length of the rope.
Thus,
$$T-mg=\dfrac{mv^2}{L}$$
Solving for $v$;
$$\dfrac{(T-mg)L}{ m}=v^2 $$
$$v=\sqrt{\dfrac{(T-mg)L}{ m}}$$
Plugging the known;
$$v=\sqrt{\dfrac{(5-[0.1\times 9.8])0.6}{ 0.1}}=\bf 4.91\;\rm m/s$$
Now we have a ball that is fired horizontally at a speed of 4.91 m/s at a height of $(y_0=2-0.6=\bf1.4\;\rm m)$ above the ground and we need to find its horizontal displacement.
The ball is now under free-fall acceleration, thus we can find the time of its trip until it touches the ground by applying the kinematic formula of vertical displacement.
$$y=y_0+v_{0y}t+\frac{1}{2}a_yt^2$$
where $y=0$ since it is the ground, $y_0=1.4$ m, $v_{0y}=0$ since it is fired horizontally, and $a_y=-g=-9.8$ m/s/s since it is under the free fall acceleration.
$$0=1.4+0-4.9 t^2$$
So that
$$t=\sqrt{\dfrac{1.4}{4.9}}=\bf 0.5345 \;\rm s$$
Now we know the time of its trip and we know that its horizontal velocity component is constant and is equal to the releasing velocity.
Thus, the distance is given by
$$x=\overbrace{x_0}^{=0}+v_{0x}t+\frac{1}{2}\overbrace{a_y}^{=0}t^2$$
$$x=vt=4.91\times0.5345 $$
$$x=\color{red}{\bf 2.62}\;\rm m$$
