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**Hint:**Recall the concept of potential energy. The potential energy of a particle is defined as the energy that is stored in an object due to its position or height relative to the mean position or from where it started moving. The object can store energy due to its position.

**Complete step by step solution:**

If a particle starts from a mean position and has started moving a small distance ‘dx’ then a small amount of work ‘dU’ is to be done. The force applied on the particle to move from the mean position is then given by:

$F = - \dfrac{{dU}}{{dx}}$---(i)

Substituting the value of ‘U’ in the above equation, we get

$\Rightarrow F = - \dfrac{d}{{dx}}(2 - 20x + 5{x^2})$

Solving the derivative, of the force

$\Rightarrow F = - ( - 20 + 10x)$

$\Rightarrow F = 20 - 10x$---(ii)

For the particle to be in equilibrium, all the forces acting on the particle should be zero. This means it can be written that

$\Rightarrow F = 0$

From equation (ii), it is clear that

$\Rightarrow 20 - 10x = 0$

$\Rightarrow 20 = 10x$

$\Rightarrow x = 2m$

It is given that the particle is released at $x = - 3$, so the amplitude at $x = - 3$ is $5m$

The maximum amplitude at the positive x-axis is then given by

$ \Rightarrow x = 2 + 5 = 7m$

**Option D is the right answer.**

**Note:**It is important to remember that the potential energy is also of two types: gravitational potential energy and elastic potential energy. If an object is placed at a height from the ground level it is said to possess gravitational potential energy. But the elastic potential energy is the energy stored in the object such that they can be compressed or stretched.

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