Answer
$25.9\;\rm cm$
Work Step by Step
Here, we have two motion stages: the first is when the granite cube slides down without friction until it reaches the bottom, and the second is when it moves horizontally at a constant velocity just before it hits the steel cube.
We need to find the speed of the granite cube just before it hits the steel cube. And then, we can find the initial height that will give it this speed.
Let's assume that the collision between the two cubes is perfectly elastic. We can use the conservation of momentum since the momentum is conserved during the short time of the collision.
Before and after the collision:
$$p_{i}=p_f$$
$$m_1v_1=m_1v_{1,fx}+m_2v_{2,fx}$$
where $v_1$ is the speed of the granite cube at the bottom of the ramp and just before it hits the steel cube.
We know, from the given, that $m_2=2m_1=2m=200\;\rm g$.
$$\color{red}{\bf\not} mv_1=\color{red}{\bf\not} m v_{1,fx}+2\color{red}{\bf\not} mv_{2,fx}$$
$$ v_1= v_{1,fx}+2v_{2,fx}$$
We also know that the final speed of the steel cube is 1.5 m/s.
Thus,
$$ v_1= v_{1,fx}+(2\times 1.5)$$
Thus,
$$ v_1-3= v_{1,fx}\tag 1$$
Now we need to find $v_1$;
We chose the positive $x$-direction of the first stage of motion of the granite cube to be down the ramp and parallel to its surface while the positive $y$-direction is perpendicular to the ramp's surface.
The ramp is frictionless which means that the only force that makes the granite cube slides down the ramp is the $x$ component of its weight.
Thus,
$$\sum F_x=mg \sin\theta =ma_x$$
Hence, its acceleration down the ramp is
$$a_x=g \sin\theta\tag 2$$
Now we can use the kinematic formula of velocity squared to find $v_1$;
$$v_1^2=v_0^2+2a_x d$$
where $d$ is the distance traveled by the cube down the ramp.
We know that the cube is released from rest, so
$$v_1^2=0^2+2a_x d$$
Plugging from (2);
$$v_1^2= 2 dg \sin\theta $$
Noting that $\sin\theta =
\dfrac{h}{d}$ where $h$ is the height from which the cube where released. So, $d =\dfrac{h}{\sin\theta}$
thus,
$$v_1^2= 2 \left(\dfrac{h}{\sin\theta}\right)g \sin\theta $$
So,
$$v_1=\sqrt{2gh}\tag 3$$
Now we can use the conservation of energy since we considered that the collision is perfectly elastic,
$$E_i=E_f$$
$$K_{i1}+K_{i2}+U_{i1}+U_{i2}=K_{f1}+K_{f2}+U_{f1}+U_{f2}$$
Whereas $1$ refers to the granite cube and $2$ for the steel cube.
We know that the steel cube was initially at rest and was on the ground level. We also know that both cubes will be on the ground level.
Thus,
$$K_{i1}+\overbrace{K_{i2}}^{=0}+U_{i1}+\overbrace{U_{i2}}^{=0}=K_{f1}+K_{f2}+\overbrace{U_{f1}}^{=0}+\overbrace{U_{f2}}^{=0}$$
$$\frac{1}{2}\color{red}{\bf\not} mv_1^2+\color{red}{\bf\not} mgy=\frac{1}{2}\color{red}{\bf\not} mv_{1f,x}^2+\frac{1}{\color{red}{\bf\not} 2}(\color{red}{\bf\not} 2\color{red}{\bf\not} m)(1.5)^2$$
$$\frac{1}{2} v_1^2+ gh=\frac{1}{2} v_{1f,x}^2+ (1.5)^2$$
Plugging from (3);
$$\frac{1}{\color{red}{\bf\not} 2}(\color{red}{\bf\not} 2gh)+ gh=\frac{1}{2} v_{1f,x}^2+ (1.5)^2$$
$$ gh=\frac{1}{2} v_{1f,x}^2+ (1.5)^2$$
Plugging from (1);
$$ gh=\frac{1}{2} (v_1-3)^2+ (1.5)^2$$
$$ gh=\frac{1}{2} (v_1^2-6v_1+9) + (1.5)^2$$
$$ gh=0.5 v_1^2-3v_1+4.5 + 2.25$$
Plugging from (3);
$$ gh=0.5 (2gh) -3(\sqrt{2gh}) + 6.75$$
Thus,
$$ 0 = -3(\sqrt{2gh}) + 6.75$$
$$ 3(\sqrt{2gh}) = 6.75 $$
$$ 2gh =\left(\dfrac{ 6.75}{3}\right)^2 $$
$$ h =\dfrac{1}{2g}\left(\dfrac{ 6.75}{3}\right)^2 =\dfrac{1}{2\times 9.8}\left(\dfrac{ 6.75}{3}\right)^2 $$
$$h=\color{red}{\bf 0.259}\;\rm m=\color{red}{\bf 25.9}\;\rm cm$$
