Convert $k$ hours to minutes, then remember that one nozzle receives only an equal fraction of the total flow.

The injector adds 3 cubic centimeters of nutrient solution per 250 cubic centimeters of water, so multiply the water amount by $\frac{3}{250}$.

Algebra is faster here because the quantities are variables. Desmos can be used to check the unit-rate calculation with convenient positive values.

Set $r=5$, $m=4$, $n=3$, and $k=2$. Enter r/m*60*k*1000/n*3/250 to calculate the nutrient amount from the situation.

Enter each answer-choice expression using the same values. Compare each resulting value with the value from the situation; only one choice matches.

The matching expression is $\frac{720rk}{mn}$, so the answer is choice D.

The system delivers water at a rate of $\frac{r}{m}$ liters per minute. In $k$ hours, there are $60k$ minutes, and one nozzle receives $\frac{1}{n}$ of the total water flow.

Convert liters to cubic centimeters by multiplying by $1{,}000$. Then use the nutrient ratio: for each cubic centimeter of water, the nutrient amount is $\frac{3}{250}$ cubic centimeter. Thus, multiply the one-nozzle water amount by $\frac{3}{250}$.

The numerical factor is $60\cdot1{,}000\cdot\frac{3}{250}=720$. Therefore, the nutrient solution added through one nozzle is $\frac{720rk}{mn}$, which is choice D.