Answer
$34.8^\circ$
Work Step by Step
The most important part of this problem is to sketch it very well so you are not confused by any given data.
The author here asks about $\theta_T$ (Target angle) in the figure below.
We can see from the geometry of the figure below that
$$\theta_T=\tan^{-1}\left[ \dfrac{3\;\rm m}{\Delta x_1+\Delta x_2} \right]\tag 1$$
So we have to find $\Delta x_1$ and $\Delta x_2$
From the geometry of the figure below, we can see that
$$\theta_2=\bf 60^\circ$$
since it is the supplementary angle.
Also, we can see that
$$\tan30^\circ=\dfrac{2}{\Delta x_1}$$
Hence,
$$\Delta x_1=\dfrac{2\;\rm m}{\tan30^\circ}=\bf 2\sqrt{3}\;\rm m\tag 2$$
Now we can find $\theta_1$ by applying Snell's law,
$$n_1\sin\theta_1=n_2\sin\theta_2$$
where $_1$ refers to water and $_2$ refers to air.
Hence,
$$\theta_1=\sin^{-1}\left[ \dfrac{n_2\sin\theta_2}{n_1}\right]$$
Plugging the known;
$$\theta_1=\sin^{-1}\left[ \dfrac{(1.0)\sin60^\circ}{1.33}\right]$$
$$\theta_1=\bf 40.63^\circ$$
Now we can see that
$$\tan(90^\circ -\theta_1)=\dfrac{1\;\rm m}{\Delta x_2}$$
Hence,
$$\Delta x_2=\dfrac{1\;\rm m}{\tan49.37^\circ}=\bf 0.858\;\rm m\tag 3$$
Now we need to plug (2) and (3) into (1),
$$\theta_T=\tan^{-1}\left[ \dfrac{3\;\rm m}{2\sqrt{3} \;\rm m+0.858\;\rm m} \right]=\color{red}{\bf 34.8^\circ}$$
So you should fire the spear at an angle of about 35$^\circ$ below the horizontal.
