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

When unpolarized light of intensity $I_0$ passes through the first polaroid, its intensity becomes $I_1 = \dfrac{I_0}{2}$.

Let the angle of the middle polaroid with the first be $\theta$. Since the third polaroid is crossed with the first, its angle with the middle polaroid is $(90^{\circ} - \theta)$.

The final transmitted intensity is:
$I = I_1 \cos^2\theta \cos^2(90^{\circ} - \theta)$
$I = \dfrac{I_0}{2} \cos^2\theta \sin^2\theta = \dfrac{I_0}{8} \sin^2(2\theta)$

For (A) $\theta = 15^{\circ}$: $I = \dfrac{I_0}{8} \sin^2(30^{\circ}) = \dfrac{I_0}{8} \times \dfrac{1}{4} = \dfrac{I_0}{32}$ (Matches II)

For (B) $\theta = 30^{\circ}$: $I = \dfrac{I_0}{8} \sin^2(60^{\circ}) = \dfrac{I_0}{8} \times \dfrac{3}{4} = \dfrac{3I_0}{32}$ (Matches IV)

For (C) $\theta = 45^{\circ}$: $I = \dfrac{I_0}{8} \sin^2(90^{\circ}) = \dfrac{I_0}{8} \times 1 = \dfrac{I_0}{8}$ (Matches III)

For (D) $\theta = 90^{\circ}$: $I = \dfrac{I_0}{8} \sin^2(180^{\circ}) = 0$ (Matches I)

Answer: (A)-(II), (B)-(IV), (C)-(III), (D)-(I)