SINCE THE RING IS ENCLOSED INSIDE THE SPHERE, SUCH THT THE CENTRE OF THE SPHERE TOUCH THE CIRCUMFERENCE OF THE RING, U CAN VISUALISE OR ASSUME A RING ON THE SPHERE COPLANER WITH THE CHARGED RING. NOW IF U DRAW THE DIAGRAN U WILL SEE THT THE PORTION OF THE RING INSIDE THE SPHERE IS SUTENDING AN ANGLE 2pie/3 rad AT THE CENTRE OF THE SPHERE.NOW SINCE THE LINEAR CHARGE DENSITY OF THE RING IS UNIFORM U CAN CALCULATE THT Q/3 CHARGE IS INSIDE THE SPHERE SINCE ONE - THIRD PORTION OF RING IS INSIDE THE SPHERE.THEREFORE THE FLUX PASSING THROUGH THE SPHERE WILL BE

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