1. Interpret the setup

   Block $A$ of mass $m_A=1\,\text{kg}$ is placed on top of block $B$ of mass $m_B=3\,\text{kg}$.

   A horizontal force $F$ is applied on block $B$.

   Coefficient of friction:

   • Between $A$ and $B$: $\mu_{AB}=0.2$
   • Between $B$ and table: $\mu_{Bt}=0.2$

   We need the maximum $F$ such that block $A$ does not slip on block $B$.

2. Condition for no slipping of block $A$ on $B$

   For block $A$ to move together with $B$, friction between $A$ and $B$ must be able to provide the needed acceleration.

   Maximum static friction between $A$ and $B$ is $f_{\max}=\mu_{AB} N = 0.2 \times (1\times 10)=2\,\text{N}$

   Since the only horizontal force on block $A$ is friction, $f = m_A a$

   For no slipping, $m_A a \le f_{\max}$ $1\cdot a \le 2$ $a \le 2\,\text{m/s}^2$

   So the maximum allowed common acceleration is $a_{\max}=2\,\text{m/s}^2$

3. Friction between block $B$ and table

   Total normal force on the table due to both blocks is $N_{table}=(m_A+m_B)g=(1+3)\times 10=40\,\text{N}$

   Hence friction between $B$ and table is $f_{table}=\mu_{Bt}N_{table}=0.2\times 40=8\,\text{N}$

4. Apply Newton's second law to the combined system $(A+B)$

   Treating both blocks as one system of mass $m_A+m_B=4\,\text{kg}$

   Net horizontal force on the system is $F - 8$

   Therefore, $F - 8 = 4a$

   For the maximum force without slipping, use $a=a_{\max}=2$: $F - 8 = 4\times 2 = 8$ $F = 16\,\text{N}$

5. Check options

   • A: $8\,\text{N}$ → too small
   • B: $16\,\text{N}$ → correct
   • C: $40\,\text{N}$ → would cause slipping
   • D: $12\,\text{N}$ → below maximum

Final Answer: $\boxed{16\,\text{N}}$ So, the correct option is B.