What is the physical meaning of magnifying power of a telescope?

I believe the reason for the confusion is that choices 3 and 4 actually refer to the same physical situation and in fact are both correct. They both are linked to the following geometric law: an object with height $s$ and distance $r$ from the observer will have an apparent angular size $\theta$ given by:

$\theta=s/r$

A telescope magnifies the apparent angular size of objects; in this case, an object of size $\theta$ is magnified to $20\times\theta$. Since $\theta$ corresponds to the above ratio, one could say that the numerator $s$ had been made 20 times larger (i.e. the object is 20 times taller). However, one could also say that the denominator $r$ was 20 times smaller (i.e. the object is 20 times nearer). Either one leads to the correct angular size.

May be interesting to note that $$\theta=s/r$$ is dependent on the small angle approximation $$ \sin(\theta)=\theta $$ which would be a good approximation in most applications of telescopes.

A object that is 20 times nearer would require re-focusing of near field optics. For terrestrial telescopes, cameras and binoculars the intuition of focusing according to r could seem to be at odds with magnification "bringing objects nearer".

Additionally an object that is 20 times nearer or 20 times taller will be 20 times brighter (- if only we had telescopes like that!). In many cases the naked eye won't notice that a magnified image is too dim, but it is relevant in photography exposure times and probably to other "measurements". So I find I would go for answer e) (none of the above) and instead say that the image of the object is 20 times larger.

Maybe the reason people can't agree on one of the options provided is because none of them are really very good.

I think probably_someone's answer is already very good, but I wanted to do some math instead of studying, so here we go:

Consider the tree with height $h=10 \mathrm{m}$ at a distance $d$. The apparent size of the object can be expressed in terms of the angle $\theta$ it occupies, that is given by $$ \begin{align} \tan \theta &= h/d \quad (\text{opposite over adjacent)} \\ \Rightarrow \theta &\simeq h/d \quad \text{(small angle approximation)} \end{align} $$ Let's call this initial angle $\theta_1$.

Then consider the two cases $h'= h \cdot 20$ or $d'=d/20$. In case the height is 20 times larger, $\theta_2$ becomes (in small angle approximation) $$ \theta_2 = h'/d = 20 \cdot \frac{h}{d} = 20 \cdot \theta_1 $$ and $\theta_3$ becomes $$ \theta_3 = h/d' = \frac{h}{\frac{d}{20}} = 20 \cdot \frac{h}{d} = 20 \cdot \theta_1 $$ and answers 3 and 4 are equivalent, as long as the small angle approximation holds. Given that the question states "...a distant tree..." I think this is a valid approximation to make.