Answer
0.0455% 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 $80^{\circ}$
We can find an expression for $I_2$:
$I_2 = I_1~cos^2~80^{\circ}$
$I_2 = (\frac{1}{2}I_0)~cos^2~80^{\circ}$
Note that the angle between $\theta_2$ and $\theta_3$ is $80^{\circ}$
We can find an expression for $I_3$:
$I_3 = I_2~cos^2~80^{\circ}$
$I_3 = (\frac{1}{2}I_0~cos^2~80^{\circ})~cos^2~80^{\circ}$
$I_3 = \frac{1}{2}I_0~cos^4~80^{\circ}$
$I_3 = 0.000455~I_0$
We can find the percentage of light that is transmitted:
$0.000455 \times 100\% = 0.0455\%$
0.0455% of the initial intensity is transmitted by the system.
