The electric field in a certain region is acting radially outward and is given by $E = Ar^2$. A charge contained in a sphere of radius 'a' centred at the origin of the field, will be given by

$4πε_0Aa^4$

The correct answer is Option (3) → $4πε_0Aa^4$

Given electric field:

$ E = A r^2 $

Using Gauss’s law:

$ \Phi_E = \oint \vec{E} \cdot d\vec{S} = \frac{Q}{\epsilon_0} $

On a spherical surface of radius $a$:

$ E = A a^2 $

So, flux:

$ \Phi_E = E \cdot 4 \pi a^2 = (A a^2)(4 \pi a^2) = 4 \pi A a^4 $

Therefore, enclosed charge:

$ Q = \epsilon_0 \Phi_E = \epsilon_0 (4 \pi A a^4) $

Charge contained = $ 4 \pi \epsilon_0 A a^4 $