Answer
$\approx 4.0\;\rm rpm$
Work Step by Step
We assume that the system is rotating without friction which means that the angular momentum of the system is constant and conserved.
$$L_i=L_f$$
We know that initially the track and the toy car were at rest. So, $L_i=0$
$$0=L_f$$
$$L_f=I_{car}\omega_{car}+I_{track}\omega_{track}=0$$
Hence,
$$ I_{car}\omega_{car}=-I_{track}\omega_{track}$$
So that the final angular velocity of the track is given by
$$\omega_{track}=\dfrac{I_{car}\omega_{car}}{-I_{track}}\tag 1$$
The negative sign indicates that the two object are rotating in opposite directions which is reasonable since the the net momentum of both is zero.
Now we need to find the moment of inertia of both of them.
$$I_{track}=I_{hoop}=M_{track}R_{track}^2$$
And since the car is at the edge of the hoop;
$$I_{car}=M_{car}R_{track}^2$$
Plug these last two results into (1);
$$\omega_{track}=\dfrac{M_{car} \color{red}{\bf\not}R_{track}^2\omega_{car}}{-M_{track} \color{red}{\bf\not}R_{track}^2} $$
$$\omega_{track}=\dfrac{M_{car} \omega_{car}}{-M_{track} } \tag 2$$
Now we need to find the angular velocity of the car where we know its velocity relative to the track is 0.75 m/s.
Thus,
$$v_{car}-v_{track}=0.75$$
where $v_{track}$ is the linear speed of any point at the end of the hoop track. Hence,
$$v_{car}=0.75+v_{track}$$
Recall that $v=\omega R$; so
$$\omega_{car} R_{track}=0.75+\omega_{track}R_{track}$$
Gividing both sides by $R$;
$$\omega_{car} =\dfrac{0.75}{R_{track}}+\omega_{track} $$
Plugging into (2);
$$\omega_{track}=\dfrac{M_{car} \left[\dfrac{0.75}{R_{track}}+\omega_{track} \right]}{-M_{track} } $$
Solving for $\omega_{track}$;
$$-M_{track}\omega_{track}= M_{car} \left[\dfrac{0.75}{R_{track}}+\omega_{track} \right] $$
$$-M_{track}\omega_{track}= M_{car} \dfrac{0.75}{R_{track}}+M_{car} \omega_{track} $$
$$- M_{car} \dfrac{0.75}{R_{track}} = M_{track}\omega_{track}+M_{car} \omega_{track} $$
$$ \dfrac{-0.75M_{car} }{R_{track}} = \left[M_{track} +M_{car} \right]\omega_{track} $$
$$ \omega_{track} = \dfrac{-0.75M_{car} }{R_{track}\left[M_{track} +M_{car} \right]} $$
Plugging the known;
$$ \omega_{track} = \dfrac{-0.75 (0.20)}{0.30\left[1+0.20\right]} = \bf 0.4167\;\rm rad/s $$
Hence, the angular velocity of the track in rpm is given by
$$ \omega_{track} = 0.4167\;\rm rad/s \left(\dfrac{1\;rev}{2\pi \;rad}\right)\left(\dfrac{60\;s}{1\;min}\right)=\color{red}{\bf 3.98 }\;\rm rpm$$
