#### Answer

We can rank the intensities of light transmitted through the second polarizer, from greatest to smallest:
$a = e \gt b \gt d \gt c$

#### Work Step by Step

Since the light is randomly polarized initially, the intensity of the light after passing through the first polarizer is $\frac{I_0}{2}$
We can use the law of Malus to determine an expression for the intensity after passing through the second polarizer.
$I = \frac{I_0}{2}~cos^2(\vert \theta_2-\theta_1\vert)$
For each situation, we can find an expression for the intensity of the light after passing through the two polarizers.
(a) $I = \frac{I_0}{2}~cos^2(30^{\circ}-0^{\circ}) = \frac{3}{8}\times I_0$
(b) $I = \frac{I_0}{2}~cos^2(30^{\circ}-30^{\circ}) = \frac{1}{2}\times I_0$
(c) $I = \frac{I_0}{2}~cos^2(90^{\circ}-0^{\circ}) = 0\times I_0$
(d) $I = \frac{I_0}{2}~cos^2(60^{\circ}-0^{\circ}) = \frac{1}{8}\times I_0$
(e) $I = \frac{I_0}{2}~cos^2(60^{\circ}-30^{\circ}) = \frac{3}{8}\times I_0$
We can rank the intensities of light transmitted through the second polarizer, from greatest to smallest:
$a = e \gt b \gt d \gt c$