Given that the horizontal range of a projectile is three times its maximum height, we need to determine the value of $ n $ in the expression for the range $\frac{n u^2}{25 g}$.

Let's start by setting up the equations for the range and maximum height of a projectile.

The formula for the range $ R $ is:

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

The formula for the maximum height $ H_{\max} $ is:

$ H_{\max} = \frac{u^2 \sin^2 \theta}{2g} $

According to the problem, the range $ R $ is three times the maximum height $ H_{\max} $:

$ R = 3H_{\max} $

Substituting the formulas for $ R $ and $ H_{\max} $ into this condition gives:

$ \frac{u^2 \sin 2\theta}{g} = 3 \cdot \frac{u^2 \sin^2 \theta}{2g} $

Simplifying this equation, we get:

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

Using the identity $ \sin 2\theta = 2 \sin \theta \cos \theta $, we can rewrite:

$ 2 \sin \theta \cos \theta = \sin 2\theta $

This gives us:

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

Dividing both sides by $\sin \theta$, assuming $\sin \theta \neq 0$, we have:

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

Rearranging the terms to solve for $\tan \theta$, we get:

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

Thus, the angle $\theta$ where $\tan \theta = \frac{4}{3}$ corresponds to $\theta = 53^\circ$.

Now, substituting back into the range formula:

$ R = \frac{u^2 \cdot 2 \cdot \frac{3}{5} \cdot \frac{4}{5}}{g} $

The calculation yields:

$ R = \frac{24 u^2}{25 g} $

Therefore, the value of $ n $ in the expression $\frac{n u^2}{25 g}$ is $ n = 24 $.