The oxidation state of $\text{Cr}$ in $[\text{Cr}({\text{NH}}_3{)}_4{\text{Cl}}_2{]}^{+}$ is:

Let the oxidation state of $\text{Cr}$ be $x$.Oxidation state of ${\text{NH}}_3$ (neutral ligand) $=0$.Oxidation state of $\text{Cl}$ (anionic ligand) $=-1$.Total charge on the complex $=+1$.Calculation:
$$x+4(0)+2(-1)=+1$$
$$x-2=+1$$
$$x=+3$$
The oxidation state of Chromium is $+3$.