Charge is uniformly distributed in a space. The net flux passing through the surface of an imaginary cube of side a in the space is “​x​. The net flux passing through the surface of an imaginary sphere of radius a in the space will be

E.S =X (given for cube)E. 6a^2 = X (S of cube = 6a^2)Then E= X/6a^2 Now X/6a^2×a^3 (E.f per unit volume)Now as given in the questions the E.f is uniformly distributed so,E.f through 4/3πr^3(which is volume of Sphere) is (4πa^3/3) ×(X/6a^2 ×a^3)=Which is equal to 4πX/18a^2Now Flux= E.f× Area(4πX/18a^2)×4πa^2(Area of sphere) =8π^2X/9 (Answer)From my point of view!😊