As weknow that, the magnifying power $m$ is the ratio of the angle $\beta$ subtended at the eye by the final image to the angle $\alpha$ subtended by object.
$$
\mathrm{m}=\frac{\beta}{\alpha}=\frac{\mathrm{h}}{\mathrm{f}_{\mathrm{e}}} \frac{\mathrm{f}_0}{\mathrm{~h}} \quad \text { (in normal adjustment) }
$$
or $m=\frac{f_0}{f_e}$
In this case, the length of the telescope tube is $f_0+f_e$,
So, $\quad \mathrm{L}=\mathrm{f}_0+\mathrm{f}_{\mathrm{e}}$
As given that, $\mathrm{f}_0=20 \mathrm{~m}$,
$$
\mathrm{f}_{\mathrm{e}}=0.02 \mathrm{~m}
$$
So, the length of the telescope tube is
So,
$$
\begin{aligned}
(\mathrm{L})=\mathrm{f}_0+\mathrm{f}_{\mathrm{e}}=\underset{\mathrm{f}}{ }=\underset{\mathrm{f}_0}{20}+(0.02)=20.02 \mathrm{~m} \\
\mathrm{~m}=\frac{\mathrm{f}_{\mathrm{c}}}{\mathrm{c}}, \\
\mathrm{m}=20 / 0.02=1000
\end{aligned}
$$
So, the final image formed is inverted and real.