I assume #log# means logarythm with base #10#. To calculate such logarythm you have to think of it as a solution to the equation:
#10^x=0.001#
So you look for a real number #x# to which you have to raise #10# to get #0.001#
#0.001=1/1000=1/10^3=10^(-3)#, so to get #0.001# you have to raise #10# to the power of #(-3)#, so this is the solution to the equation and answer to the task: