The correct choice is (c) 23
For explanation: We have M = 3 x 5 x 7 = 105; M/3 = 35; M/5 = 21; M/7 = 15.
The set of linear congruences
35 x b1 = 1 (mod 3); 21 x b2 = 1 (mod 5); 15 x b3 = 1 (mod 7)
has the solutions b1 = 2; b2 = 1; b3 = 1. Then,
x = 2 x 2 c 35 + 3 x 1 x 21 + 2 x 1 x 15 = 233 (mod 105) = 23.