Answer
$10.13^\circ$
Work Step by Step
First of all, we need to draw a sketch that represents the situation clearly, as you see below.
We need to find the component of the initial velocity of the boat relative to Earth. We chose left to be our positive $x$-direction.
$$v_{x,bE}=v_{x,bw}+v_{x,wE}=v_0\cos\theta+(-2)$$
whereas $b\rightarrow$ boat, $w\rightarrow$ water, and $E\rightarrow$ Earth.
From the right triangle below, we can see that $\theta$ when the boat is directly directed toward the kid is given by
$$\tan\theta=\dfrac{200}{1500}\Rightarrow \theta=7.59^\circ$$
Thus,
$$v_{x,bE}= 8\cos7.59^\circ -2=\bf 5.93\;\rm m/s\tag 1$$
$$v_{y,bE}=v_{y,bw}+v_{y,wE}=v_0\sin\theta+0$$
$$v_{y,bE}=8\sin7.59^\circ =\bf 1.06\;\rm m/s\tag 2$$
Thus, the angle that the pilot had to leave the shore to go directly to the child is given by
$$\theta_{bE}=\tan^{-1}\left[\dfrac{v_{y,bE}}{v_{x,bE}}\right]=\tan^{-1}\left[\dfrac{1.06 }{ 5.93}\right]$$
$$\theta_{bE}=\color{red}{\bf 10.13^\circ}$$
