1. **State the problem:** We need to find how many times greater the number of viruses is compared to the number of phytoplankton in a 1-liter sea water sample. 2. **Given data:** - Number of viruses = $3 \times 10^9$ - Number of phytoplankton = $1 \times 10^6$ 3. **Formula used:** To find how many times greater one quantity is than another, divide the larger quantity by the smaller quantity: $$\text{Ratio} = \frac{\text{Number of viruses}}{\text{Number of phytoplankton}}$$ 4. **Calculate the ratio:** $$\frac{3 \times 10^9}{1 \times 10^6} = 3 \times 10^{9-6} = 3 \times 10^3$$ 5. **Simplify:** $$3 \times 10^3 = 3 \times 1000 = 3000$$ 6. **Interpretation:** The number of viruses is 3000 times greater than the number of phytoplankton. 7. **Rounding:** Since 3000 is already a whole number, the nearest whole number is 3000. **Final answer:** The number of viruses is approximately **3000 times greater** than the number of phytoplankton.