Answer
$x = 1.9~m$
Work Step by Step
We can find the angle from the normal that the light ray from the near side of the meter stick makes with the water surface:
$tan~\theta_n = \frac{1.0~m}{3.0~m}$
$\theta_n = tan^{-1}~(\frac{1.0~m}{3.0~m})$
$\theta_n = 18.435^{\circ}$
We can find the angle this light ray makes with the normal when it emerges from the water:
$n_a~sin~\theta_1 = n_w~sin~\theta_n$
$sin~\theta_1 = \frac{n_w~sin~\theta_n}{n_a}$
$\theta_1 = sin^{-1}~(\frac{n_w~sin~\theta_n}{n_a})$
$\theta_1 = sin^{-1}~(\frac{1.33~sin~18.435^{\circ}}{1.00})$
$\theta_1 = 24.87^{\circ}$
We can find the apparent position of the near side of the meter stick:
$\frac{x_n}{3.0~m} = tan~24.87^{\circ}$
$x_n = (3.0~m)~(tan~24.87^{\circ})$
$x_n = 1.39~m$
We can find the angle from the normal that the light ray from the far side of the meter stick makes with the water surface:
$tan~\theta_f = \frac{2.0~m}{3.0~m}$
$\theta_f = tan^{-1}~(\frac{2.0~m}{3.0~m})$
$\theta_f = 33.695^{\circ}$
We can find the angle this light ray makes with the normal when it emerges from the water:
$n_a~sin~\theta_2 = n_w~sin~\theta_f$
$sin~\theta_2 = \frac{n_w~sin~\theta_f}{n_a}$
$\theta_2 = sin^{-1}~(\frac{n_w~sin~\theta_f}{n_a})$
$\theta_2 = sin^{-1}~(\frac{1.33~sin~33.695^{\circ}}{1.00})$
$\theta_2 = 47.54^{\circ}$
We can find the apparent position of the far side of the meter stick:
$\frac{x_f}{3.0~m} = tan~47.54^{\circ}$
$x_f = (3.0~m)~(tan~47.54^{\circ})$
$x_f = 3.28~m$
We can find the apparent length of the meter stick:
$x = x_f-x_n$
$x = (3.28~m)-(1.39~m)$
$x = 1.9~m$
