Understand the arrangement

The positive integers are written row-wise so that:

• 1st row has $1$ number,
• 2nd row has $2$ numbers,
• 3rd row has $3$ numbers,
• and so on.

So, after $n$ rows, the total numbers written are $1+2+3+\cdots+n = \frac{n(n+1)}{2}.$

Find the row containing 5310

We need the smallest integer $n$ such that $\frac{n(n+1)}{2} \ge 5310.$

Also, the previous row must satisfy $\frac{(n-1)n}{2} < 5310.$

Solve the inequality

Check nearby values:

$\frac{102\cdot 103}{2} = 51\cdot 103 = 5253$ $\frac{103\cdot 104}{2} = 103\cdot 52 = 5356$

Since $5253 < 5310 \le 5356,$ the number $5310$ lies after the end of row $102$ and within row $103$.

Conclusion

Therefore, the number $5310$ is in the $\boxed{103^{\text{rd}}\text{ row}}.$