The cups are arranged in a $3 \times 3$ matrix, with 6 cups containing coffee and 3 cups containing tea. The goal is to find the number of ways they can be arranged such that each row contains at least one cup of coffee.

1. Total Arrangements: The total number of ways to arrange the 9 cups (6 coffee and 3 tea) without any restrictions is calculated using the formula for permutations of multiset:

Total arrangements = $\frac{9!}{6! \times 3!}$ This gives the total number of ways to arrange the cups in the grid.

2. Removing Invalid Cases: The condition is that each row must contain at least one cup of coffee. To account for cases where a row contains only tea, we need to subtract the number of arrangements where one row contains all 3 cups of tea.

There are 3 rows, so we can have 3 different scenarios where all 3 cups of tea are in a single row. For each such case, the remaining 6 coffee cups can be arranged in the remaining positions.

This gives 3 cases to subtract from the total.

3. Final Answer: The required number of arrangements is:

Required arrangements = Total arrangements - 3 = 84 - 3 = 81

Hence, the correct answer is 81.