When dealing with three polaroids, the intensity after the second polaroid will be given by Malus' law as $I_0 \cos^2 \theta$, where $\theta$ is the angle between the pass axes of the first two polaroids. However, as the third polaroid is orthogonal to the first, no light from the first polaroid passes through, only light from the second polaroid. So the final intensity is also modulated by a $\sin^2 \theta$ term (as the second and third polaroids are orthogonal).

So if we set up the equation for the final intensity $I_{\text{net}}$:

$$I_{\text{net}} = I_0 \cos^2 \theta \sin^2 \theta$$

And we substitute the given values $I_{\text{net}} = 3 \, \text{W/m}^2$ and $I_0 = \frac{32 \, \text{W/m}^2}{2} = 16 \, \text{W/m}^2$:

$$3 = 16 \cos^2 \theta \sin^2 \theta$$

This simplifies to:

$$\frac{3}{16} = \sin^2 \theta \cos^2 \theta = \left(\frac{1}{2} \sin 2\theta\right)^2$$

Taking the square root of both sides gives:

$$\frac{\sqrt{3}}{2} = \left|\sin 2\theta\right|$$

The solutions for this are $\theta = 30^\circ$ and $\theta = 60^\circ$.

So, the angle between the pass axes of the first two polaroids is either $30^\circ$ or $60^\circ$.