Answer
$4.03\;\rm m/s$
Work Step by Step
First, we need to draw a pictorial representation of our problem, as shown in the first figure below. And we need to draw the force diagram of the car, as seen in the second figure below.
Now we need to apply Newton's second law;
$$\sum F_r=T\sin\theta=ma_r=m\dfrac{v^2}{R}$$
$$ T\sin\theta =m\dfrac{v^2}{R} $$
where $R$ is the radius of the circular path which is here given by $(R_{disk}+d)$, as seen below, and $t$ is the periodic time.
$$ T\sin\theta =m\dfrac{v^2}{R_{disk}+d} $$
Now we need to find $d$ which is given by
$$\sin\theta=\dfrac{d}{L_{chain}}\rightarrow d=L_{chain}\sin\theta$$
So that,
$$ T\sin\theta =\dfrac{mv^2}{R_{disk}+(L_{chain}\sin\theta)} \tag 1$$
$$\sum F_y=T\cos\theta-mg=ma_y=m(0)=0$$
Thus,
$$T=\dfrac{mg}{\cos\theta}$$
Plugging into (1);
$$ \dfrac{mg}{\cos\theta}\sin\theta =\dfrac{mv^2}{R_{disk}+(L_{chain}\sin\theta)} $$
$$ \color{red}{\bf\not}mg\tan\theta =\dfrac{ \color{red}{\bf\not}mv^2}{R_{disk}+(L_{chain}\sin\theta)} $$
$$ g\tan\theta =\dfrac{ v^2}{R_{disk}+(L_{chain}\sin\theta)} $$
Therefore,
$$v =\sqrt{ g\tan\theta\left[ R_{disk}+(L_{chain}\sin\theta)\right]}$$
Plugging the known;
$$v =\sqrt{ 9.8\tan20^\circ\left[ 2.5+6\sin20^\circ \right]}$$
$$v =\color{red}{\bf 4.03}\;\rm m/s$$
