A long string with charge per unit length $\mathrm{\lambda }$ passes through of a cube of length I The maximum flux through the cube will be
 

 

See fig. The maximum length of the string which can be fitted in the cube is along its body diagonal. Hence maximum flux passes its this position. Charge enclosed $=\sqrt{3}\cdot \mathrm{l\lambda }$ 

$\therefore \mathrm{\Phi }=\frac{\sqrt{3}\mathrm{\lambda l}}{{\mathrm{\epsilon }}_0}$