Easiest way of calculating this is: 600/9= 66.66667.

So we know its in 67th box. then 66*9= 594 – which is the last (biggest) number in the 66th box. so then we just count 6 more numbers with the ascending pattern shown in the 3 boxes above and you come to r3c2

the alternative answer i suggested was to watch the multiples of 6 and 10.
all multiples of 6 will show on the bottom line, alternating between the middle column in a grid, and the first and third on the next.
but the great deal is to realize that all multiples of 10 will be in the same cell as its “starting number”
for instance: 5 will be in the same position as 50, 6 in the same position as 60. and the same thing will happen with the multiples of 100: 300 in the same position as 3, so 600 in the same position as 6.

Bilbao, as soon as I posted my answer I knew there was something else that was going on with problem after I went back an re-read it.

The sum of the digits is the position in the block. Didn’t see that at all… Neat.

Notice numbers divisile by 3 are in the bottom row. Numbers divisible by 9 are in the bottom row as well as the last column.

So 600 is divisible by 3, so we know it is in the bottom row in the 200th column.
66 x 9 = 594. So 594 is bottom row, last column of the 66th block. So 6 more numbers would be in the middle column, on the bottom row of the 67th block.

600 = 594 + 6
594 is one of the multiples of 9 so will appear at the position of 9. So in the next box 600 will appear at the position of 6. ( Middle column, bottom row )

if the remainder is 1, number fall in 1st box (col 1, row 1)
if the remainder is 2, number fall in 2nd box (col 1, row 2)
and so on
if the remainder is 0, number fall in last box (col 3, row 3)

600 will fall in 2nd column, third line