Four identical blocks of length L are arranged one over the other as shown. The maximum distance of the uppermost block from the edge of the lowermost block is x such that no block tumbles. Then x is

For the blocks to not topple, 

The COM of the blocks above should lie inside the bottom block's length. 

For the top 2 blocks :

X₁ = l/2

For the next 2 blocks, 

x₂ = l/4 

For the next 2 blocks, 

x₃ = l/6

x = x₁ + x₂ + x₃ $\left[\therefore {\mathrm{x}}_3=\frac{1}{2\mathrm{n}}\right]$
$=\frac{1}{2}+\frac{1}{2\left(2\right)}+\frac{1}{2\left(3\right)}\Rightarrow \mathrm{x}=\frac{11\mathrm{l}}{12}$