Unpolarized light of intensity I_0 is incident on a system of three...

Intensity of light after passing through the first polaroid is $I_1 = \dfrac{I_0}{2}$.

Let the angle between the transmission axes of the first and second polaroids be $\theta$. The intensity after the second polaroid is $I_2 = I_1 \cos^2\theta$.

Since the first and third polaroids are crossed, the angle between the second and third polaroids is $(90^{\circ} - \theta)$.

The final intensity after the third polaroid is:
$I_3 = I_2 \cos^2(90^{\circ} - \theta) = I_1 \cos^2\theta \sin^2\theta$

Using the identity $\cos\theta \sin\theta = \dfrac{\sin(2\theta)}{2}$, we get:
$I_3 = \dfrac{I_1}{4} \sin^2(2\theta) = \dfrac{I_0}{8} \sin^2(2\theta)$

Given that $I_3 = \dfrac{I_0}{32}$, we have:
$\dfrac{I_0}{8} \sin^2(2\theta) = \dfrac{I_0}{32}$
$\sin^2(2\theta) = \dfrac{1}{4}$
$\sin(2\theta) = \dfrac{1}{2}$

This gives $2\theta = 30^{\circ}$, which means $\theta = 15^{\circ}$.

Answer: $15^{\circ}$