Answer
44% of the initial intensity is transmitted by the system.
Work Step by Step
On the graph, we can see that no light is transmitted when $\theta_2 = 160^{\circ}$
Then $\theta_1 = 70^{\circ}$
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.
$I_1 = \frac{1}{2}I_0$
It is given that $\theta_2 = 90^{\circ}$
Note that the angle between $\theta_1$ and $\theta_2$ is $20^{\circ}$
We can find an expression for $I_2$:
$I_2 = I_1~cos^2~20^{\circ}$
$I_2 = (\frac{1}{2}I_0)~cos^2~20^{\circ}$
$I_2 = 0.44~I_0$
We can find the percentage of light that is transmitted:
$0.44 \times 100\% = 44\%$
44% of the initial intensity is transmitted by the system.
