A charge Q is uniformly distributed over a rod of length l.consider a hypothetical cube of edge l with the center of the cube at one end of the rod.find minimum possible flux of the electric field through thr entire surface.

answer Q/(2*"apsylum not")

Is this a doubt or you are just posting it for everyone?
If it is a doubt, you can use gauss law
total flux = Q(enclosed)/(apsylum not)
Min Q(enclosed) for this case is Q/2 by keeping the rod perpendicular to any surface. Hence the result.

sir it was a doubt

i did not get it keeping the rod perpendicular how eill it chenge q enclosed

If you had taken any other orientation, the length of rod enclosed would have been more. ( Like if you had taken the rod to be passing from the center through the edge of the cube, the length enclosed would have been L/(root 2), hence the charge enclosed would have been more.)

Equivalently, you can see that the shortest distance between the center of the cube, and the 'outside' of the cube would have to be L/2. Hence minimum amount of the rod with one end at the center (and charge) which could be enclosed would be L/2 (Q/2).