1. Interpret the first statement

The given microscopic lens alone is used like a simple microscope. For a simple microscope, when the final image is formed at the near point $D=25\,\text{cm}$, the magnifying power is

$M=1+\frac{D}{f}$

Given magnification $M=6$,

$1+\frac{25}{f_o}=6$ $\frac{25}{f_o}=5$ $f_o=5\,\text{cm}$

So the focal length of the objective (the given lens) is

$f_o=5\,\text{cm}$

2. Microscope with objective + eyepiece

Now the same lens is used as the objective along with an eyepiece and tube length

$L=0.6\,\text{m}=60\,\text{cm}$

The final image is at infinity, so for a compound microscope the magnifying power is

$M=\left(\frac{L}{f_o}\right)\left(\frac{D}{f_e}\right)$

where

• $f_o=5\,\text{cm}$ is the objective focal length,
• $f_e$ is the eyepiece focal length,
• $D=25\,\text{cm}$.

3. Use the condition on total magnification

The new total magnification is double the earlier magnification. Earlier magnification was $6$, so new magnification is

$M=2\times 6=12$

Thus,

$\left(\frac{60}{5}\right)\left(\frac{25}{f_e}\right)=12$

$12\cdot \frac{25}{f_e}=12$

$\frac{25}{f_e}=1$

$f_e=25\,\text{cm}$

4. Final answer

$\boxed{25}$

This matches the stored correct answer.