Let's distribute that #color(white)(^)4#:
#((2a^2)^4)/((3a^6b^2)^4)#
Now we can continue to distribute
#(2^4a^(2xx4))/(3^4a^(6xx4)b^(2xx4))#
#(16a^8)/(81a^24b^8)#
Now, is there anything we can divide out?
#16/81 xx a^8/a^24 xx 1/b^8#
To divide with exponents, we actually subtract. So #a^8/a^24# is the same thing as #a^(8-24)#, or #a^-16#. Nothing else can be simplified:
#16/81 xx a^-16/1 xx 1/b^8#
We're almost done, but first we need to do something about that #a^-16#. To change a negative expoent, we need to swith it's position. If it is negative in the denominator, moving it to the numerator makes it positive, and same for when it is negative in the numerator, like in our case. So, now we have #1/a^16#
#16/81 xx 1/a^16 xx 1/b^8#
Let's combine these back together, and then we're done
#16/(81a^16b^8)#
