Perfect squares are the middle term of odd rows. $(17{)}^2=289$ should be middle term of ${17}^{th}$ row. Option (a).

Alternate Solution:

There is 1 odd number in the $1^{st}$ row.

There are 2 odd numbers in the $2^{nd}$ row.

There are 3 odd numbers in the $3^{rd}$ row. ...

Hence this can be considered as $a\sum n$ question where, n is the number of odd numbers.

As 289 will be $\frac{289}{2}=\frac{144}{{145}^{th}}$ term in the sequence, find $a\sum n$ value which is close to the number. which is equal to 153 and 17 numbers before 153 will be in the ${17}^{th}$ row. Hence the answer is 17.