The target row sum is $9 \cdot 10/2 \cdot 3 = 15$. We first note that the small numbers $1$, $2$, and $3$ must be in separate rows since if two of them were in the same row, the sum would be too small: we would get at most $2 + 3 + 9 = 14$. In a similar fashion, the large numbers $7$, $8$, and $9$ must be in separate rows. It follows that each row has one small and one large number.

Multiples of $3$ cannot all be in the same row (since $3+6+9 > 15$), nor can a row have exactly two multiples of $3$ (the sum would not be divisible by $3$). Thus, each row has exactly one multiple of $3$. It follows that each row has numbers in all mod-$3$ congruence classes.

Given these constraints and ignoring the order of the elements, there are just two possibilities for the row containing $1$ ($\{1, 5, 9\}$ or $\{1, 6, 8\}$) and for the row containing $2$ ($\{2, 4, 9\}$ or $\{2, 6, 7\}$). Because of overlapping elements, each possible row containing $1$ is compatible with only one possible row containing $2$. Once these two rows are chosen, the final row is also known, so there are only two ways of allocating the elements to rows.

These three rows can be put in any order, e.g. first the row with $1$, then the one with $2$, etc., giving $3! = 6$ possible re-orderings of rows. Internal to each row, we can re-order the elements in $3!$ ways. Thus, the total number of row-balanced squares is $6^4 \cdot 2 = 2^5 \cdot 3^4$.