Three blocks are kept as shown in figure. Acceleration of 20kg block with respect to ground is 

Maximum value of friction between 10kg and 20kg is

${\left(f_1\right)}_{\max }=0.5\times 10\times 10=50 \mathrm{N}$

Maximum value of friction between 20kg and 30kg is

${\left(f_2\right)}_{\text{max }}=(0.25)(10+20)\left(10\right)=75 \mathrm{N}$

Now ,assume that 20kg and 30kg move as a single block with 10kg block. So, let us first calculate the requirement of $f_1$ for this

$$\begin{gathered} 100-f_1=10a \\ {f}_1=50a \end{gathered}$$

On solving these two equations, we get

$f_1=83.33 \mathrm{N}$

Since, it is greater than ${\left(f_1\right)}_{\text{mux }},s_0$ there is slip between 10kg and other two blocks and $50 \mathrm{N}$ will act here.

Now let us check whether there is slip between $20 \mathrm{kg}$ and 30kg or not. For this we will have to calculate requirement of $f_2$ for no slip condition.

And $50-f_2=20a$

On solving these two equations, we get

and

$$\begin{gathered} f_2=30 \mathrm{N} \\ a=1 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Since, $f_2$ is less than ${\left(f_2\right)}_{\max }$, so there is no slip between $20 \mathrm{kg}$ and 30kg and both move together with same acceleration of $1 \mathrm{m}/{\mathrm{s}}^2$.