Vernier constant $= \large\frac{1}{N}$$S = \large\frac{1}{10}$$ = 0.1mm$

The instrument has a positive error which is $e = NC = 7 \times 0.1mm = 0.07cm$

Vernier scale reading $= 4 \times 0.01 = 0.04cm$ (since the fourth vernier division coincides with a scale division)

Therefore, the observed reading = $3.1 + 0.04 = 3.14cm$

The true reading, adjusting for error is $ = 3.14 - 0.07 = 3.07cm$