Answer
Yes.
Work Step by Step
To maintain the shown group of bricks on the table without falling from the edge we need to make the net center of the mass never exceeds the edge of the table.
So the maximum distance $d$ occurs when the center of the mass of the bricks group is exactly of the edge. Otherwise, when the center of the mass of the group exceeds the edge the bricks will move clockwise and fall down.
So we will work from the first brick to the lowest one.
The center of the mass of brick A could be above the right edge of brick B and stays in equilibrium.
Now we need to find the center of the mass of the two upper bricks A and B so we can put their center of mass just above the right edge of brick C.
We chose the origin of the first two bricks to be the left edge of brick B.
$$x_{cm,(A+B)}=\dfrac{m_Ax_A+m_Bx_B}{m_A+m_B}$$
where $m_A=m_B=m_C=m_D=m$; and $x_A=L$ while $x_B=\frac{1}{2}L$
$$x_{cm,(A+B)}=\dfrac{m L+m\frac{L}{2}}{2m}=\color{blue}{\dfrac{3L}{4}}$$
Now we need to put the center of mass of this combination (brick A+brick B) above the right edge of brick C. This means that only one-fourth of brick B will be on the air, as you see in the second figure below.
Finding the center of mass of this new combination A+B+C.
We chose the origin to be the left edge of brick C.
$$x_{cm,(A+B+C)}=\dfrac{2mx_{cm,(A+B)}+m_Cx_C}{2m+m_c}$$
$$x_{cm,(A+B+C)}=\dfrac{2mL+m \dfrac{L}{2}}{2m+m}=\color{blue}{\dfrac{5L}{6}}$$
Now we need to put the center of mass of this combination (brick A+brick B+brike C) above the right edge of brick D. This means that only one-sixth of brick C will be on the air, as you see in the third figure below.
We chose the origin to be the left edge of brick D.
$$x_{cm,(A+B+C+D)}=\dfrac{3mx_{cm,(A+B+C)}+m_Dx_D}{3m+m_c}$$
$$x_{cm,(A+B+C)}=\dfrac{3mL+m \dfrac{L}{2}}{3m+m}=\color{blue}{\dfrac{7L}{8}}$$
Now we need to put the center of mass of this combination (brick A+brick B+brike C+brick D) above the edge of the table. This means that only one-eighth of brick D will be on the air, as you see in the fourth figure below.
Now we need to find $d$ which is the distance from the edge of the table to the right edge of brick A. Noting that the center of the mass of the four bricks is just at the edge of the table.
We can see from the last figure below that
$$d=\dfrac{L}{2}+\dfrac{L}{4}+\dfrac{L}{6}+\dfrac{L}{8}=\color{red}{\dfrac{25L}{24}}$$
Therefore, yes, no part of the upper brick A is over the table since $d\gt L$.
