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**Hint:**The mass of the block $A$ is determined by using the equation of the motion of the block $B$ and the buoyancy force act on the block $B$. By equating both the equation of the block $B$, the mass of the block $A$ can be determined.

**Useful formula:**

The buoyancy force is given by,

$F = V \times \rho \times g$

Where, $F$ is the buoyancy force, $V$ volume of the block, $\rho $ is the density of the liquid and $g$ is the acceleration due to gravity.

**Complete step by step solution:**

Given that,

The mass of the block $B$ is, ${m_B} = 3\,m$.

The density of the liquid is, $\rho $.

The density of the block $B$ is, $2\rho $.

Now,

The block $B$ moves down as it be out of liquid, so the equation of the motion is given by,

$3mg - Mg = \left( {3m + M} \right)a\,.............\left( 1 \right)$

The above equation is defined as, the force of the block $B$ and the force of the block $A$ are moved in opposite directions, so both are subtracted, is equal to the total force due to block $A$ and block $B$.

The density of the block $B$ is, $2\rho $.

The mass of the block $B$ is, ${m_B} = 3\,m$.

So, the volume of the block $B$ is, $V = \dfrac{{3m}}{{2\rho }}$

The buoyancy force is given by,

$ \Rightarrow V \times \rho \times g$

By substituting the volume of the block $B$, then

$ \Rightarrow \dfrac{{3m}}{{2\rho }} \times \rho \times g$

On further simplification, then

$ \Rightarrow \dfrac{{3mg}}{2}$

Now, the equation of motion when the block $B$ is inside the liquid and moving up, then

$Mg + \dfrac{3}{2}mg - 3mg = \left( {3m + M} \right)a\,..............\left( 2 \right)$

The buoyancy force pushes the block $B$ upwards, so it is added.

By equation the equation (1) and equation (2), then

\[3mg - Mg = Mg + \dfrac{3}{2}mg - 3mg\]

By arranging the above equation, then

$3mg + 3mg - \dfrac{3}{2}mg = Mg + Mg$

By adding the above equation, then

$6mg - \dfrac{3}{2}mg = 2Mg$

On cross multiplying the terms in LHS, then

$\dfrac{9}{2}mg = 2Mg$

By rearranging the terms, then

$Mg = \dfrac{9}{4}mg$

By cancelling the terms on both sides, then

$M = \dfrac{9}{4}m$

Thus, the above equation shows the mass of the block $A$.

**Hence, the option (D) is correct.**

**Note:**The block $B$ moves in two conditions, so the two equations of the motion of the block $B$ is written and these two equations of the motion of the block $B$ is equated, then the mass of the block $A$ is determined in terms of $m$.

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