Answer
$0.03125~~$ of the initial intensity is transmitted by the system.
Work Step by Step
Let $I_0$ be the original intensity of the light.
Since the light is initially unpolarized, half the intensity will be transmitted through the first polarizing sheet.
$I_1 = \frac{1}{2}I_0$
Note that the angle between $\theta_1$ and $\theta_2$ is $60^{\circ}$
We can find an expression for $I_2$:
$I_2 = I_1~cos^2~60^{\circ}$
$I_2 = \frac{1}{2}I_0~cos^2~60^{\circ}$
Note that the angle between $\theta_2$ and $\theta_3$ is $60^{\circ}$
We can find an expression for $I_3$:
$I_3 = I_2~cos^2~60^{\circ}$
$I_3 = \frac{1}{2}I_0~cos^2~60^{\circ}~cos^2~60^{\circ}$
$I_3 = \frac{1}{2}I_0~cos^4~60^{\circ}$
$I_3 = 0.03125~I_0$
$0.03125~~$ of the initial intensity is transmitted by the system.
