Answer
$0.84~~$ of the light intensity received before the glasses were put on now reaches the sunbather’s eyes.
Work Step by Step
Note that intensity is proportional to the square of the electric field magnitude.
That is: $~~I \propto E^2$
Therefore:
$I_y \propto E_y^2$
$I_x \propto E_x^2 = (2.3~E_y)^2 = 5.29~E_y^2$
We can find the fraction of light intensity that remains when the vertical component is eliminated:
$\frac{I_x}{I_x+I_y} = \frac{E_x^2}{5.29~E_y^2+E_y^2} = \frac{5.29}{6.29} = 0.84$
$0.84~~$ of the light intensity received before the glasses were put on now reaches the sunbather’s eyes.
