lui $13, 0x0060
ori $13, $13, 0x5000
lw  $12, 0xC($13)

Alternate Sequence

The above ori instruction "fills in" the lower 16 bits of register $13 by doing the following:

$13 after lui           :  0000 0000 0110 0000 0000 0000 0000 0000
zero-extended imm. op.  :  0000 0000 0000 0000 0101 0000 0000 0000
result of bitwise OR    :  0000 0000 0110 0000 0101 0000 0000 0000

Other sequences of instructions also will work: Because the "upper half" of an address is 16 bits and the offset of the lw instruction is 16 bits, the two in combination can address any byte of memory.

The problem was to load $12 with the word at address 0x0060500C. Here is another way to do it: Split the address into halves: 0x0060 and 0x500C. Load the top half into $13 and use the bottom half as the offset.

lui $13, 0x0060
lw  $12, 0x500C($13)

Choose whichever method works best in the context of the rest of the program.

Since the ori instruction is used often with the destination register as one of the operands, there is a shorthand instruction in assembly language.

ori  $d,const  # assembles to the same machine
               # instruction as ori $d,$d,const

ori  $10,$10,0x00C4

ori  $10,$0, 0x00C4