1. For an astronomical telescope in normal adjustment:

   • Length of tube: $L = f_o + f_e$
   • Magnifying power: $M = \frac{f_o}{f_e}$

   where $f_o$ is the focal length of objective and $f_e$ is the focal length of eyepiece.

2. Given: $L = 60\text{ cm}, \quad M = 5$

   So, $\frac{f_o}{f_e} = 5 \Rightarrow f_o = 5f_e$

3. Substitute into tube length equation: $f_o + f_e = 60$ $5f_e + f_e = 60$ $6f_e = 60$ $f_e = 10\text{ cm}$

4. Hence, the focal length of the eyepiece is $\boxed{10\text{ cm}}$

5. Checking options:

   • A: $40$ cm $\to$ incorrect
   • B: $10$ cm $\to$ correct
   • C: $30$ cm $\to$ incorrect
   • D: $20$ cm $\to$ incorrect

Therefore, the correct option is B.