Answer
$q' = 7.45 nC $
Work Step by Step
Let the charge on the inner conductor be –q. Immediately adjacent to it is the induced charge q' . Since the electric field is less by a factor $\frac{1}{k}$ than the field when no dielectric is present, then $-q+q' = -\frac{q}{k} $
Thus,
$q' = \frac{k-1}{k} q = \frac{23.5 - 1}{23.5} (7.79 nC)$
$q' = 7.45 nC $
