Answer
The intensity of light that is transmitted by the system is $~~19~W/m^2$
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 $20^{\circ}$
We can find an expression for $I_2$:
$I_2 = I_1~cos^2~20^{\circ}$
$I_2 = (\frac{1}{2}I_0)~cos^2~20^{\circ}$
$I_2 = 0.4415~I_0$
We can find the intensity of light that is transmitted by the system:
$I = (0.4415)(43~W/m^2)$
$I = 19~W/m^2$
The intensity of light that is transmitted by the system is $~~19~W/m^2$
