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 55,246. To provide you the exact information about the admission cutoff would be tough as JEE Main is being conducted twice by NTA for the first time but with this rank chances seems less to get a seat in NITs.

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

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!