From the data given in the question,
Pitch of the screw gauge=0.5mm.
Total divisions on the circular scalen=50.
Therefore, the least count of the screw gauge is LC=npitch=500.5=0.01mm
Now, MSR=2.5mm, CSR=45
Diameter reading =MSR+LC×CSR− zero error
⇒d=2.5+0.45−(−0.03)
⇒d=2.98mm