The answer is #"0.237 mM"#.
So, you start with #"11.2 mg"# of glucose and #"7.45 mL"# of water. In order to determine the molarity of this solution, you must find the number of moles of glucose you have
#11.2 * 10^(-3)"g" * ("1 mole")/("180 g") = 0.0622 * 10^(-3)"moles"#
This makes the concentration of your parent solution
#C = n/V = (0.0622 * 10^(-3)"moles")/(7.45 * 10^(-3)L) = "0.00835 M"#
The next step is to determine how many moles of glucose the aliquot will contain
#n = C * V = 0.00835("mol")/L * 17.0 * 10^(-6)"L" = 0.142 * 10^(-6)#
The volume of the new solution will be
#V_("solution") = 17.0mu"L" + 583mu"L" = 600.0mu"L"#
Therefore, the concentration will be
#C = n/V = (0.142 * 10^(-6)"moles")/(600.0 * 10^(-6)"L") = 0.000237"moles"/"L"#, or
#C = 0.237"mmol"/"L" = "0.237 mM"#
