The work done against the retarding force is indeed equal to the loss in kinetic energy.

The force acting on the particle due to retardation is given by $F = ma = -2mx$.

When we integrate this force over the displacement from $0$ to $x$, we get:

$$\Delta KE = W = \int F \cdot dx = \int (-2mx) \, dx = -mx^2$$

The negative sign indicates that this is a loss of kinetic energy.

The problem states that the loss in kinetic energy is also given by $\left(\frac{10}{x}\right)^{-n}$ J. Therefore, we have:

$$-mx^2 = \left(\frac{10}{x}\right)^{-n}$$

Because this is a loss of kinetic energy, we should consider the absolute value. Hence,

$$mx^2 = \left(\frac{10}{x}\right)^{-n}$$

Substituting the given mass $m = 10 \, \text{g} = 0.01 \, \text{kg}$, we get:

$$0.01x^2 = \left(\frac{10}{x}\right)^{-n}$$

This simplifies to:

$$x^2 = \left(\frac{10}{x}\right)^{-n}$$

Comparing the two sides, we can see that $n = 2$.