Suppose C1 and C2 are two circles with the same center O.
And AC is a chord touching C1 at the point D
let’s join OD.
So, $OD\bot AC$
$AD=DC=4cm$ [perpendicular line OD bisects the chord]
Thus, in right angled ∆AOD,
$O{{A}^{2}}=A{{D}^{2}}+D{{O}^{2}}$ [Derived from Pythagoras theorem]
$D{{O}^{2}}={{5}^{2}}-{{4}^{2}}$
$\Rightarrow 25-16=9$
$DO=3cm$
So, radius of the inner circle OD is equals to $3cm$.