The total number of elements in the first n rows is:
\[ S = 1 + 2 + 3 + \dots + T_n = \frac{n(n+1)}{2}. \]
To find the row containing 5310, solve:
\[ \frac{n(n+1)}{2} = 5310. \]
Start testing values:
\[ n = 100, \quad T_n = \frac{100 \cdot 101}{2} = 5050. \]
\[ n = 101, \quad T_n = \frac{101 \cdot 102}{2} = 5151. \]
\[ n = 102, \quad T_n = \frac{102 \cdot 103}{2} = 5253. \]
\[ n = 103, \quad T_n = \frac{103 \cdot 104}{2} = 5356. \]
Since 5310 lies between 5253 and 5356, it is in the 103rd row.
Final Answer: 103.