Here the [Cr(NH_{3}) _{4}Cl_{2}]^{+} 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**