Here the [Cr(NH3) 4Cl2]+ has a positive charge.
NH3 has zero charge.
Cl has one -ve charge.
Let the oxidation number of Cr be x.
So simple mathematical equation for charge on compound will be -
x + 4*0 + 2*(-1) = +1
Solving we get
x = +3
So, Oxidation number of Cr in [Cr(NH3) 4Cl2]^+ is +3