The total charge induced on the inner surface of the dielectric shell is given by:
\[ Q_{\text{induced}} = -Q \left( 1 - \frac{1}{\varepsilon_r} \right) \]
Substituting \( Q = -9C \) and \( \varepsilon_r = 9 \):
\[ Q_{\text{induced}} = -(-9) \left( 1 - \frac{1}{9} \right) \]
\[ Q_{\text{induced}} = 9 \left( 1 - \frac{1}{9} \right) \]
\[ Q_{\text{induced}} = 9 \left( \frac{9 - 1}{9} \right) \]
\[ Q_{\text{induced}} = 9 \times \frac{8}{9} \]
\[ Q_{\text{induced}} = 8C \]
Since the calculation rounds off to two decimal places, we write:
\[ Q_{\text{induced}} \approx 8.00C \]
Thus, the total charge induced on the inner surface of the dielectric shell is \( 8.00C \).

At a particular temperature T, Planck's energy density of black body radiation in terms of frequency is \(\rho_T(\nu) = 8 \times 10^{-18} \text{ J/m}^3 \text{ Hz}^{-1}\) at \(\nu = 3 \times 10^{14}\) Hz. Then Planck's energy density \(\rho_T(\lambda)\) at the corresponding wavelength (\(\lambda\)) has the value \rule{1cm}{0.15mm} \(\times 10^2 \text{ J/m}^4\). (in integer)
[Speed of light \(c = 3 \times 10^8\) m/s]
(Note: The unit for \(\rho_T(\nu)\) in the original problem was given as J/m³, which is dimensionally incorrect for a spectral density. The correct unit J/(m³·Hz) or J·s/m³ is used here for the solution.)