Identify the forces and contacts

Two blocks are pressed horizontally against a vertical wall.

• Weight of block $A = 20\,\text{N}$
• Weight of block $B = 100\,\text{N}$
• Coefficient of friction between $A$ and $B$: $\mu_{AB}=0.1$
• Coefficient of friction between $B$ and wall: $\mu_{BW}=0.15$

Since the system is at rest, friction must balance the downward tendency due to weights.

Friction on block $A$

Block $A$ is supported vertically only by friction from block $B$.

Therefore, for block $A$ in vertical equilibrium, $f_{AB}=20\,\text{N}$ acting upward on $A$.

By Newton's third law, block $A$ exerts an equal and opposite friction on block $B$ of magnitude $20\,\text{N}$ downward.

Vertical forces on block $B$

Downward forces on block $B$ are:

• Its own weight: $100\,\text{N}$
• Friction due to block $A$: $20\,\text{N}$

So total downward force on $B$ is $100+20=120\,\text{N}$

Friction by the wall on block $B$

To keep block $B$ in equilibrium, the wall must provide an upward frictional force equal to this total downward force: $f_{wall\to B}=120\,\text{N}$

Check with options

The frictional force applied by the wall on block $B$ is $\boxed{120\,\text{N}}$

So the correct option is A.

Comparison with stored answer

Stored correct answer: A

Our derived answer: A

Hence, they agree.