To solve this problem, we will use Gauss's Law, which relates the electric field to the charge enclosed by a Gaussian surface.$\\$Gauss's Law is given by:$\\ \oint E \cdot dA = \frac{Q_{enc}}{\varepsilon_0} \\$Where:$\\$• $E$ is the electric field.$\\$• $dA$ is the differential area element.$\\$• $Q_{enc}$ is the charge enclosed by the Gaussian surface.$\\$• $\varepsilon_0$ is the permittivity of free space.$\\$Given that the electric field $E = A r,$ where $A$ is a constant and $r$ is the radial distance from the origin.$\\$Consider a spherical Gaussian surface of radius $a$ centered at the origin.$\\$The electric field is radial and has the same magnitude at every point on the surface, so:$\\ \oint E \cdot dA = E \cdot 4\pi a^2 \\$Substitute $E = A a$ into the equation:$\\ A a \cdot 4\pi a^2 = \frac{Q_{enc}}{\varepsilon_0} \\$Simplify the left side:$\\ 4\pi A a^3 = \frac{Q_{enc}}{\varepsilon_0} \\$Solve for $Q_{enc}: \\ Q_{enc} = 4\pi \varepsilon_0 A a^3 \\$Therefore, the charge contained in the sphere of radius $a$ is $4\pi \varepsilon_0 A a^3. \\$This corresponds to Option 2.