It would be difficult to provide you the exact information about the qualifying score or Admission cutoff as JEE Main is being conducted twice by NTA for the first time. As a total No. of 1,45,386 candidates had appeared for January Session, Using the Formula (100-your score)* 1,45,386 / 100- Your Expected rank will be 65,423. With this rank the chances are less to get a seat in NITs.

You can check the previous year cutoff to take an Idea about its cutoff.

https://engineering.careers360.com/articles/jee-main-cutoff-for-b-arch-b-planning

NOTE: This is only a prediction. NTA will merge the scores obtained in first and second attempt by candidates of JEE Main Paper 2 2019 for the compilation and preparation of overall merit/rank list. The rank will be released on 30th April, 2019.

Good Luck!