Correct Answer - C
the charge inside the cube due to each line is `lambdaL`.Flux due to horizontal wire through the faces `AEDH`,`BCGF` is zero as the electric field produced will be parallel to these faces.
But the total flux due to this line is `(lambdaL)/(epsilon_(0))` as it is equally distributed among the remaining four faces .Similarly flux due to vertical coire through the faces `DHGC,AEFB` is zero.
,.Flux through `CDHG=O+(lambdaL)/(4epsilon_(0))=(lambdaL)/(4epsilon_(0))`
:.Flux through `AEHD=O+(lambdaL)/(4epsilon_(0))=(lambdaL)/(4epsilon_(0))`
Flux through `ABCD=(lambdaL)/(4epsilon_(0))=(lambdaL)/(4epsilon_(0))=(lambdaL)/(2epsilon_(0))`