First things first, we have to write out what is given to us.
Given
[E] = 4 nM
K_m = 2.8 muM
V_m = (60 muM)/min
Next, we need to use two formulas to figure this problem out
color(white)(aaaaaaaaaaa)color(red)((k_(cat))/K_m color(white)(aaaaaaaa) k_(cat) = V_m/[[E]]
We are going to figure out our k_(cat) or the turnover number of the enzyme, but first, our answer is supposed to be represented in muM so we will first convert 4nM -> muM.
(4 cancel(nM))/(1) * (1*10^-9 M)/(1 cancel(nM)) = (4 * 10^-9 cancel(M))/(1) * (1 muM)/(1 * 10^-6 cancel(M)) = darr
color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)(4 * 10^-3 muM)
Plugin
color(red)(k_(cat) = V_m/[[E]]) -> [(60 cancel(muM))/min]/[(4*10^-3cancel(muM))/1] = (15000)/min
Now that we have our k_(cat), we will use it to find out the "catalytic efficiency", which basically tells us how good the enzyme is at catalyzing the reaction at low substrate concentration.
color(red)((k_(cat))/K_m) = [(15000)/min]/[(2.8muM)/1] = (5357.14)/(min*muM)
Take this answer and convert min -> s
(5357.14)/(cancelmin*muM) * (1 cancelmin)/(60 s) = color(blue)((89)/(s*muM)
Answer: 89/(s*muM)
