Let the positive integers be written in the form:

If the $$k^{th}$$ row contains exactly $$k$$ numbers for every natural number $$k$$, then the row in which the number 5310 will be, is ________

You’re placing numbers in rows where row (k) has exactly (k) numbers. That means the last number in row (k) is the triangular number:
$$T_k=\frac{k(k+1)}{2}$$

We want to find which row contains 5310, so solve:
$$\frac{k(k+1)}{2}\ge5310$$

Multiply both sides by 2:
$$k(k+1)\ge10620$$

This gives the quadratic inequality:
$$k^2+k-10620\ge0$$
$$k=\frac{-1\pm\sqrt{42481}}{2}$$

$$\sqrt{42481}\approx206.1$$

$$k\approx\frac{-1+206.1}{2}\approx102.55$$

$$So(k\approx103).$$

Now check:

$$(T_{102}=5253)$$

$$(T_{103}=5356)$$

Since (5310) lies between these, it is in row:

103