Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
In general, to design a telescope we need to use an objective lens with a long focal length and an eye lens with a shorter focal length.
And we know that the focal length is inversely proportional to the refractive power.
$$f=\dfrac{1}{P}\tag 1$$
Hence, we can use the one that has a greater refractive power as an eye lens. And the other one as an objective lens.
Thus, $\rm+4.5\; D$ is the lens that should be used as the eyepiece, while the $\rm+3.0\; D$ is the lens that should be used as the objective lens.
$$\color{blue}{\bf [b]}$$
We know that the magnification of a telescope is given by
$$M=-\dfrac{f_{\rm obj}}{f_{\rm eye}}$$
And from (1),
$$M=-\dfrac{P_{\rm eye}}{P_{\rm obj}}$$
Plug the known;
$$M=-\dfrac{+4.5\;\rm D}{+3.0\;\rm D}=\color{red}{\bf 1.5}$$
We can ignore the negative sign since the telescope magnification is an absolute value.
$$\color{blue}{\bf [c]}$$
The minimum length of the telescope is given by the sum of the two focal lengths of the two lenses.
This minimum length is actually the distance between the two lenses.
Thus,
$$d=L=f_{\rm eye}+f_{\rm obj}$$
From (1);
$$d =\dfrac{1}{P_{\rm eye}}+\dfrac{1}{P_{\rm obj}}$$
$$d =\dfrac{1}{4.5\;\rm D}+\dfrac{1}{3\;\rm D}=0.56\;\rm m$$
$$d=\color{red}{\bf 56}\;\rm cm$$
