A student measured the length of a rod and wrote it as 3 . 50cm ....

As per the given data the vernier has a least count of $0 . 01cm$, which is consistent with the information given in the question.

The measured value of the length of the rod$= 3 . 50cm$

So, the least count of the measuring instrument must be

$0 . 01cm = 0 . 1mm$

For the vernier scale

$10MSD = 1cm = 10mm$

$\to 1MSD = 1mm$

also $9MSD = 10VSD$

$∴$~Least count (L.C.)

$= 1MSD - 1VSD$

$= ( 1 - 0 . 9 ) mm$

$= 0 . 1mm$