**Hint** We should know that a block diagram is defined as the diagram of a system where the principal parts or we can say that the functions are represented by the blocks connected by the lines that will give us an idea about the relationships of the blocks. Using this principal, we can solve the question.

**Complete step by step answer**

We know that it is given that g = 10 $m/{s^2}$.

The block diagram of 5 kg block is given as:

a = 3m/$s^2$

Let us consider that ${m_1}$ is the mass and the a be acceleration.

${N_1}$ is the normal reaction on ${m_1}$by the mass ${m_2}$

The equations are given as:

${{N}_{1}}-{{m}_{1}}g={{m}_{1}}a........(i)$

${\Rightarrow {N_1} = {m_1}g + {m_1}a}$

${N_1}$ denotes the normal reaction on the ${m_2}$by ${m_1}$.

${N_2}$ denotes the normal reaction on the ${m_2}$by floor.

The equations are given as:

${{N_2} - {N_1} - {m_2}g = {m_2}a.......(ii)}$

$\Rightarrow {N_2} = {N_1} + {m_2}g + {m_2}a$

From equation (ii) and equation (i):

${N_2} = {m_1}g + {m_1}a + {m_2}g + {m_2}a$

$\Rightarrow {N_2} = 5 \times 10 + 5 \times 3 + 10 \times 10 + 10 \times 3$

$\Rightarrow {N_2} = 50 + 15 + 100 + 30$

$\Rightarrow {N_2} = 195N$

Therefore, the force exerted on 10 kg by floor is 195 N.

**The correct answer is option C. **

**Note** We should know that a force which is acting perpendicular to the two surfaces which are in contact with each other. If the weight is the only vertical force that is acting on an object lying or that is moving on a horizontal surface, the normal reaction force is equal in magnitude, but is in the opposite direction to the weight.