Hi Saumitra,

The total number of candidates who appeared for MHT CET 2019 from the PCB group was 1,22,264 from all over India.

Therefore, according to your percentile, your rank would be-

1,22,264 * 0.0105 = 1,284 (Approximately)

Based on the MHT CET 2018 cutoff ranks of ICT, Mumbai for General Open Home State candidates, you can basically get all the branches apart from the Chemical Engineering branch whose cutoff rank was 807.

You can also check out the MHT CET 2018 cutoff for all the colleges from the given link-

https://engineering.careers360.com/articles/mht-cet-cutoff

Hope this helps you.

All the best!!!