Here is the situation
We need to find $\dfrac{ d \alpha }{d t }$
In vectorial notation, we write the speed of the drone as
$$ V = (20 \cos 30, 20 \sin 30) = (10 \sqrt{3}, 10) $$
Let's say $t$ seconds have elapsed and we have that the horizontal component of drone distance is
$$ 10 \sqrt{3} t $$
and therefore
$$ \tan \alpha = \frac{2}{10 \sqrt{3} t } $$
or
$$ \alpha = \tan^{-1} (1/10\sqrt{3}t) $$
AND THUS
$$ \frac{d \alpha}{d t } = - \frac{1}{10 \sqrt{3} t^2} \cdot \frac{1}{1 + (1/10\sqrt{3}t)^2 } $$
Is this correct?
First of all, the way you understand the term "downrange" seems to be wrong. It's actually a horizontal distance. Not vertical. Here's what Wikipedia says about it:
Secondly, nowhere in the problem does it say that the drone is 2 meters downrange from the tip of your head. It says from you! So, we have to assume that, as John Wayland Bales aptly mentioned in his comment, all measurements are with respect to a coordinate system centered at the ground level.
Here's how the problem can be represented graphically:
$x$, $y$ and $\theta$ are quantities that change with time. Let's find the relationships between them:
$$ \sin{\frac{\pi}{6}}=\frac{y}{x}\implies y=\frac{x}{2},\\ \tan{\theta}=\frac{y}{2}\implies \tan{\theta}=\frac{x}{4}\implies \theta=\tan^{-1}{\left(\frac{x}{4}\right)}. $$ Now, let's take that last result with the angle theta and differentiate both sides of that equation with respect to time (don't forget that $\theta$ and $x$ are functions of time):
$$\frac{d\theta}{dt}=\frac{1}{1+\left(\frac{x}{4}\right)^2}\cdot\frac{1}{4}\cdot\frac{dx}{dt}=\frac{4}{16 + x^2}\cdot\frac{dx}{dt}$$
We know what $\frac{dx}{dt}$ is. But what is $x$ equal to? When $y$ is 10 meters, $x$ is going to be 20 meters: $$y=\frac{x}{2}\implies x=2y=2\cdot 10=20\ m.$$ Thus: $$\frac{d\theta}{dt}=\frac{4}{16 + 20^2}\cdot 20=\frac{5}{26}\ rad/s.$$