Explanation

The number XXYXX = 11011X + 100Y = 11(1001X) + 100Y. Divisibility by 11 requires the alternating digit sum (X - X + Y - X + X) = Y to be a multiple of 11, so Y = 0. Divisibility by 3 requires the digit sum 4X + Y = 4X to be divisible by 3, so X is 3, 6 or 9 (X != 0 as the leading digit). That gives exactly 3 such numbers. Hence option B.