The initial velocity for your enzyme will be #"0.417 nmol L"^(-1)"s"^(-1)#.
Once again, use the MIchaelis-Menten equation to solve for #V_0#. Keep in mind, however, that you must convert #mu"mol L"^(-1)"s"^(-1)#, the units used for #V_"max"#, to #"nmol L"^(-1)"s"^(-1)#, the units required for #V_0#.
#500cancel(mu"mol")"L"^(-1)"s"^(-1) * (10^(-3)"nmol")/(1cancel(mu"mol")) = 500 * 10^(-3)"nmol L"^(-1)"s"^(-1)#
So, the initial velocity will be equal to
#V_0 = (V_"max" * [S])/(K_m + [S])#
#V_0 = (500 * 10^(-3)"nmol L"^(-1)"s"^(-1) * 5.00cancel("mmol L"^(-1)))/((1.00 + 5.00)cancel("mmol L"^(-1))#
#V_0 = 5/6 * 500 * 10^(-3)"nmol L"^(-1)"s"^(-1) = "0.417 nmol L"^(-1)"s"^(-1)#
SIDE NOTE Since #V_"max"# is given with one sig fig, while all other variables are given with three, I'll leave the answer rounded to three sig figs.
If, instead of 500 you'd have 500. for #V_"max"#, then the answer should indeed have three sig figs; if not, it should only have one.
