Number of sets $=\dfrac{\text{cache size}}{\text{(size of a block * No. of blocks in a set)}}$
$=\dfrac{128 * 64}{(64 * 4)}\text{ (4 way set associative means 4 blocks in a set)}$
$= 32.$
So, number of index (LINE) bits $= 5$ and number of WORD bits $= 6$ since cache block (line) size is $64.$
So, number of TAG bits $= 20 - 6 - 5 = 9.$
Answer is (D) choice