A red marble can be drawn from Box I afterthe two rounds of transfers in the following four ways:
Case 1: R$\to$R$\to$R
Probability = $(\frac{5}{11}\times \frac{5}{110}\times \frac{5}{11})$
Case 2: R$\to$B$\to$R
Probability = $(\frac{5}{11}\times \frac{5}{110}\times \frac{4}{11})$
Case 3: B$\to$B$\to$R
Probability = $(\frac{6}{11}\times \frac{6}{110}\times \frac{5}{11})$
Case 4: B$\to$R$\to$R
Probability = $(\frac{6}{11}\times \frac{6}{110}\times \frac{6}{11})$
Thus resulting probability = sum of probabilities of individual cases
=$(\frac{5}{11}\times \frac{5}{110}\times \frac{5}{11})$ +$(\frac{5}{11}\times \frac{5}{110}\times \frac{4}{11})$ +$(\frac{6}{11}\times \frac{6}{110}\times \frac{5}{11})$ +$(\frac{6}{11}\times \frac{6}{110}\times \frac{6}{11})$ = $\frac{621}{1210}$
Therefore, answer option (c) is the correct answer choice.