The reading of voltmeter in circuit shown is

Let R₁ be the effective resistance in parallel combination of voltmeter and 60​Ω
∴ R₁= 60x40/(60+40)
        ​= 24 Ω​

  ​  R_eff= R₁+40
          ​= 24+40
          =64Ω​
E=IR_eff
 ​6= I x 64
=>I= 3/32 A
Let voltage across the 60Ω​ and voltmeter be V
V=IR₁
   =3/32 x 24
   =2.25V

Since the voltmeter and resistor is connected in parallel they are at same potential
V=2.25V