A particle is projected with velocity $$u$$ so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as $$\dfrac{nu^2}{25g}$$, where value of n is: (Given 'g' is the acceleration due to gravity).

The standard formulae for a projectile launched with speed $$u$$ at an angle $$\theta$$ above the horizontal are:

Horizontal range:
$$R = \frac{u^{2}\sin 2\theta}{g}$$

Maximum height:
$$H = \frac{u^{2}\sin^{2}\theta}{2g}$$

The condition given in the problem is that the range is three times the maximum height:

$$R = 3H$$

Substitute the expressions for $$R$$ and $$H$$:

$$\frac{u^{2}\sin 2\theta}{g} = 3\left(\frac{u^{2}\sin^{2}\theta}{2g}\right)$$

Cancel the common factors $$u^{2}$$ and $$g$$:

$$\sin 2\theta = \frac{3}{2}\sin^{2}\theta$$ $$-(1)$$

Use the trigonometric identity $$\sin 2\theta = 2\sin\theta\cos\theta$$ and substitute it into $$(1)$$:

$$2\sin\theta\cos\theta = \frac{3}{2}\sin^{2}\theta$$

Divide both sides by $$\sin\theta$$ (since $$\theta \neq 0^{\circ},90^{\circ}$$ for a projectile with non-zero range):

$$2\cos\theta = \frac{3}{2}\sin\theta$$

Rearrange to get the cotangent of the launch angle:

$$\frac{\cos\theta}{\sin\theta} = \frac{3}{4} \quad\Longrightarrow\quad \cot\theta = \frac{3}{4}$$

Hence
$$\tan\theta = \frac{4}{3}$$

From the right-triangle definition, this gives

$$\sin\theta = \frac{4}{5},\qquad \cos\theta = \frac{3}{5}$$

Now compute $$\sin 2\theta$$ using $$\sin 2\theta = 2\sin\theta\cos\theta$$:

$$\sin 2\theta = 2\left(\frac{4}{5}\right)\left(\frac{3}{5}\right) = \frac{24}{25}$$

Finally, substitute $$\sin 2\theta$$ into the range formula:

$$R = \frac{u^{2}\sin 2\theta}{g} = \frac{u^{2}}{g}\left(\frac{24}{25}\right) = \frac{24\,u^{2}}{25g}$$

The question writes the range as $$\dfrac{nu^{2}}{25g}$$. Comparing gives $$n = 24$$.

Therefore, the correct option is Option D (24).