By Gauss's law for \(\vec D\), inside the dielectric \(\vec D=\dfrac{Q}{4\pi r^2}\hat r\), so
\[
\vec E=\frac{Q}{4\pi\varepsilon_0\varepsilon_r r^2}\hat r,\qquad
\vec P=\varepsilon_0(\varepsilon_r-1)\vec E=\frac{\varepsilon_r-1}{\varepsilon_r}\,\frac{Q}{4\pi r^2}\hat r .
\]
On the inner surface (\(r=a=0.1\) m) the outward normal of the dielectric is \(-\hat r\), so \(\sigma_b=\vec P\cdot(-\hat r)\) and
\[
Q_{b,\text{inner}}=4\pi a^2\sigma_b=-\frac{\varepsilon_r-1}{\varepsilon_r}\,Q=-\frac{8}{9}\times(-9\ \mathrm C)=\boxed{8.00\ \mathrm C}
\]
(independent of the radii; the outer surface carries \(-8\) C).
Answer: 8.00 (range 7.90 to 8.10)
