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(C) Consider a Gaussian surface as a sphere of radius $r$ such that $a < r < b$.According to Gauss's law,$\oint_{S} \vec{E} \cdot d\vec{S} = \frac{q_{

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$\frac{Q}{2\pi a^2}$

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(C) Consider a Gaussian surface as a sphere of radius $r$ such that $a According to Gauss's law,$\oint_{S} \vec{E} \cdot d\vec{S} = \frac{q_{enclosed}}{\epsilon_0}$.Here,$q_{enclosed} = Q + \int_{a}^{r} \rho(r') \cdot 4\pi r'^2 dr'$.Given $\rho(r') = \frac{A}{r'}$,we have $q_{enclosed} = Q + \int_{a}^{r} \frac{A}{r'} \cdot 4\pi r'^2 dr' = Q + 4\pi A \int_{a}^{r} r' dr' = Q + 4\pi A \left[ \frac{r'^2}{2} \right]_{a}^{r} = Q + 2\pi A (r^2 - a^2)$.Applying Gauss's law: $E \cdot 4\pi r^2 = \frac{Q + 2\pi A (r^2 - a^2)}{\epsilon_0}$.$E = \frac{1}{4\pi \epsilon_0 r^2} [Q - 2\pi A a^2 + 2\pi A r^2] = \frac{1}{4\pi \epsilon_0} [\frac{Q - 2\pi A a^2}{r^2} + 2\pi A]$.For the electric field $E$ to be constant (independent of $r$),the coefficient of $\frac{1}{r^2}$ must be zero.Therefore,$Q - 2\pi A a^2 = 0$.$A = \frac{Q}{2\pi a^2}$.

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