Answer
The maximum value of $~~a_3~~$ is $~~\frac{L}{6}$
Work Step by Step
In part (a), we found that $a_1 = \frac{L}{2}$
In part (b), we found that $a_2 = \frac{L}{4}$
The center of mass of the top three bricks must be directly above part of the fourth brick. If the center of mass of the top three bricks is above only air, the top three bricks will fall.
We can find the center of mass of the top three bricks as a distance from the right side of the diagram:
$\frac{(m)(L/2)+(m)(L)+(m)(5L/4)}{3m} = \frac{11mL/4}{3m} = \frac{11L}{12}$
We can find the maximum value of $a_3$:
$\frac{L}{2}+\frac{L}{4}+a_3 = \frac{11L}{12}$
$a_3 = \frac{11L}{12}-\frac{L}{2}-\frac{L}{4}$
$a_3 = \frac{L}{6}$
The maximum value of $~~a_3~~$ is $~~\frac{L}{6}$
