Answer
9.5 volts
Work Step by Step
The capacitor is disconnected from the battery, so the charge on it stays the same. However, the capacitance is increased by a factor of the dielectric constant, $\kappa$.
Let $Q_i$ and $V_i$ denote the initial charge and voltage on the capacitor, and let $Q_f$ and $V_f$ denote the final charge and voltage. Equation 17–7, Q=CV, relates the charges and voltages to the capacitance.
$$Q=C_iV_i=C_fV_f$$
$$V_f=V_i\frac{C_i}{C_f}= V_i\frac{C_i}{\kappa C_i}$$
$$V_f= (21.0V)\frac{1}{2.2}=9.5V$$