Four harmonics of equal frequencies and equal intensities I_0 have...

Let the amplitude of each harmonic wave be $A_0$. The intensity of each wave is $I_0 = k A_0^2$.

The waves can be represented as vectors (phasors) with magnitudes $A_0$ and angles $0, \dfrac{\pi}{3}, \dfrac{2\pi}{3}, \pi$.

The x-component of the resultant amplitude is:
$A_x = A_0 \cos(0) + A_0 \cos\left(\dfrac{\pi}{3}\right) + A_0 \cos\left(\dfrac{2\pi}{3}\right) + A_0 \cos(\pi)$
$A_x = A_0 \left( 1 + \dfrac{1}{2} - \dfrac{1}{2} - 1 \right) = 0$

The y-component of the resultant amplitude is:
$A_y = A_0 \sin(0) + A_0 \sin\left(\dfrac{\pi}{3}\right) + A_0 \sin\left(\dfrac{2\pi}{3}\right) + A_0 \sin(\pi)$
$A_y = A_0 \left( 0 + \dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{3}}{2} + 0 \right) = \sqrt{3} A_0$

The resultant amplitude $A_R$ is given by:
$A_R = \sqrt{A_x^2 + A_y^2} = \sqrt{0 + (\sqrt{3} A_0)^2} = \sqrt{3} A_0$

The resultant intensity $I_R$ is proportional to the square of the resultant amplitude:
$I_R = k A_R^2 = k (\sqrt{3} A_0)^2 = 3 k A_0^2 = 3 I_0$

Comparing this with $n I_0$, we get $n = 3$.

Answer: $3$