Two full turns of the circular scale of a screw gauge having 50...

Distance cover in $1$ full turn will be~$\dfrac{1mm}{2} = 0 . 5mm$.

Least count of screw gauge,

$LC = \dfrac{Distancecoverin1fullturn}{numberofdivisionincircularscale} = \dfrac{0 . 5mm}{50} = 0 . 01mm$

Reading of the screw gauge,$R = mainscalereading + Circularscalereading \times LC$

$R = 4mm + 37 \times 0 . 01mm = 4 . 37mm$

Reading without zero error,~$R^{'} = R - zeroerror = 4 . 37mm - \left(- 0 . 02mm\right) = 4 . 39mm$