A solution holds an enzyme at 2.5 nmol/L…

Step 1. $V_{\max}$. $\dfrac{150}{60} = 2.5$ µmol/(L s).

Step 2. $k_{cat}$. The enzyme is 2.5 nmol/L, which is 0.0025 µmol/L. So $k_{cat} = \dfrac{2.5}{0.0025} = 1000$ per second.

Step 3. The efficiency. $K_m = 0.50$ mmol/L $= 5.0 \times 10^{-4}$ mol/L. So $\dfrac{k_{cat}}{K_m} = \dfrac{1000}{5.0 \times 10^{-4}} = 2.0 \times 10^{6}$ L/(mol s).