Answer
The point of second reflection is on side 2.
Work Step by Step
Let $\theta_2$ be the refracted angle inside the block.
We can use Snell's law to find $\theta_2$:
$n_2~sin~\theta_2 = n_1~sin~\theta_1$
$sin~\theta_2 = \frac{n_1~sin~\theta_1}{n_2}$
$\theta_2 = sin^{-1}~(\frac{n_1~sin~\theta_1}{n_2})$
$\theta_2 = sin^{-1}~(\frac{1.00~sin~40^{\circ}}{1.56})$
$\theta_2 = sin^{-1}~(0.41204)$
$\theta_2 = 24.3^{\circ}$
If the ray travels a horizontal distance of $W = 3.00~cm$, we can find the vertical distance $y$ that the ray travels:
$\frac{y}{W} = tan~\theta_2$
$y = W~tan~\theta_2$
$y = (3.00~cm)~tan~24.3^{\circ}$
$y = 1.36~cm$
Since $y \lt H$, then the point of first reflection is on side 3.
The reflected angle is equal to the incident angle on side 3.
Since $2y \gt H$, then the reflected ray meets side 2 before it is able to cross the horizontal distance of $W = 3.00~cm$ back to side 1.
The point of second reflection is on side 2.
