TO SOLVE THIS U HAVE TO ASSUME THT THE CHARGE GIVEN TO OUTER CONDUCTOR WILL BE DISTRIBUTED ON THE INNER FACE AND OUTER FACE OF THE SHELL. NOW SUPPOSE Q1 CHARGE COMES ON INNER FACE OF THE SHELL.THUS AN INDUCED CHARGE -Q1 WILL BE INDUCED ON THE OUTER FACE OF THE INNER SHELL.NOW LET THE RADIUS OF  INNER SPHERES BE R3 AND R1 AND R2 BE EXTERNAL AND INTERNAL RADIUS OF OUTER SPHERE. RESPECTIVELY.SINCE INNER SPHERE IS EARTHED . ITS POTENTIAL WILL BE ZERO(ALWAYS TAKE THE POTENTIAL OF THE GROUNDED CONDUCTOR TO BE ZERO AND SOLVE).

                               (1/4pie0)[((Q-Q1/R1) + (Q1/R2) + (-Q1/R3)] = 0

SOLVING THIS U WILL GET

                                   Q1 = Q[ (R2R3) / R1R2 +R2R3 - R1R3]

NOW IF THE THICKNESS OF THE OUTER SHELL BE NEGLIGIBLE JUST PUT R1 = R2 IN THE ABOVE EQUATION .

PLZ... RATE IF SATISFIED WITH THE ANSWER.