a) Let $p_1, p_2, ..., p_7$ be the products of the elements of the rows $1, 2, ..., 7$. We have $p_1 = p_7$, $p_2 = p_6$ and $p_3 = p_5$, so $49! = (p_1 p_2 p_3)^2 \cdot p_4$, (1).

Note that $49! = 2^{46} \cdot 3^{22} \cdot 5^{10} \cdot 7^8 \cdot 11^4 \cdot 13^3 \cdot 17^2 \cdot 19^2 \cdot 23^2 \cdot 29 \cdot 31 \cdot 37 \cdot 41 \cdot 43 \cdot 47$. From (1) we infer that the primes which have odd exponents in the prime factorization of $49!$ must appear in the prime factorization of one of the numbers situated on the $4$-th row, thus $13 \cdot 29 \cdot 31 \cdot 37 \cdot 41 \cdot 43 \cdot 47 \mid p_4$. The product of any two of the primes $13, 29, 31, 37, 41, 43$ and $47$ exceeds $49$, therefore each of these primes divides exactly one of the $7$ numbers situated on the $4$-th row.

The number $\frac{49!}{p_4}$ is a perfect square, hence if $p_4 > 13 \cdot 29 \cdot 31 \cdot 37 \cdot 41 \cdot 43 \cdot 47$, we have that $p_4 \ge 2^2 \cdot 13 \cdot 29 \cdot 31 \cdot 37 \cdot 41 \cdot 43 \cdot 47$, thus one of the numbers written in the cells of the table will be at least equal to $\min\{2^2 \cdot 13, 2 \cdot 29\} > 49$, false.

Therefore, the $4$-th row contains the numbers $13, 29, 31, 37, 41, 43, 47$, whose sum is the prime $241$.