(N/A) The block is at rest on the floor. Its free-body diagram shows two forces on the block: the force of gravitational attraction by the earth equal to $2 \times 10 = 20\; N$,and the normal force $R$ of the floor on the block. By the First Law,the net force on the block must be zero,i.e.,$R = 20\; N$. Using the third law,the action of the block (i.e.,the force exerted on the floor by the block) is equal to $20\; N$ and directed vertically downwards.
$(b)$ The system (block $+$ cylinder) accelerates downwards with $0.1\; m/s^2$. The free-body diagram of the system shows two forces on the system: the force of gravity due to the earth $(27\; kg \times 10\; m/s^2 = 270\; N)$; and the normal force $R'$ by the floor. Note that the free-body diagram of the system does not show the internal forces between the block and the cylinder. Applying the second law to the system:
$270 - R' = 27 \times 0.1\; N$
$R' = 270 - 2.7 = 267.3\; N$
By the third law,the action of the system on the floor is equal to $267.3\; N$ vertically downward.
Action-reaction pairs:
For $(a)$: $(i)$ The force of gravity $(20\; N)$ on the block by the earth (action); the force of gravity on the earth by the block (reaction) equal to $20\; N$ directed upwards. $(ii)$ The force on the floor by the block (action); the force on the block by the floor (reaction).
For $(b)$: $(i)$ The force of gravity $(270\; N)$ on the system by the earth (action); the force of gravity on the earth by the system (reaction),equal to $270\; N$ directed upwards. $(ii)$ The force on the floor by the system (action); the force on the system by the floor (reaction). In addition,for $(b)$,the force on the block by the cylinder and the force on the cylinder by the block also constitute an action-reaction pair.