An unpolarized light of certain intensity passes through a...

Let the intensity of the incident unpolarized light be $I_0$.

After passing through the first polarizer, the intensity becomes $I_1 = \dfrac{I_0}{2}$.

Without the third polarizer, the light passes directly from the first polarizer (at $30^{\circ}$) to the second polarizer (at $90^{\circ}$). The angle between their transmission axes is $90^{\circ} - 30^{\circ} = 60^{\circ}$.

Using Malus's law, the output intensity without the third polarizer is:
$I_{without} = I_1 \cos^2(60^{\circ}) = \dfrac{I_0}{2} \times \left(\dfrac{1}{2}\right)^2 = \dfrac{I_0}{8}$

When the third polarizer (at $60^{\circ}$) is placed between them, the light first passes from the first polarizer to the third polarizer. The angle between their axes is $60^{\circ} - 30^{\circ} = 30^{\circ}$.

The intensity after the third polarizer is:
$I_3 = I_1 \cos^2(30^{\circ}) = \dfrac{I_0}{2} \times \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{3I_0}{8}$

The light then passes from the third polarizer to the second polarizer. The angle between their axes is $90^{\circ} - 60^{\circ} = 30^{\circ}$.

The final output intensity with the third polarizer is:
$I_{with} = I_3 \cos^2(30^{\circ}) = \dfrac{3I_0}{8} \times \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{9I_0}{32}$

The ratio of the output intensities with and without the third polarizer is:
$\text{Ratio} = \dfrac{I_{with}}{I_{without}} = \dfrac{\dfrac{9I_0}{32}}{\dfrac{I_0}{8}} = \dfrac{9}{4}$

Answer: $9/4$