In normal adjustment, for a refracting telescope, the distance...

Let ${f}_{0}$ be the focal length of objective and ${f}_{e}$ is the focal length of eyepiece.

Given here, ${f}_{0} + {f}_{e} = 30$.

Angular magnification of telescope is given by $m = \dfrac{{f}_{0}}{{f}_{e}}$.

So, $2 = \dfrac{{f}_{0}}{{f}_{e}} ⇒ {f}_{0} = 2{f}_{e}$

Then, we have ${f}_{0} + \dfrac{{f}_{0}}{2} = 30 ⇒ \dfrac{3{f}_{0}}{2} = 30$

Or, ${f}_{0} = 20cm$.