How to find oxidation state of Co and Cr in [Co(NH₃)₄(NO₂)₂][Cr(NH₃)₃(NO₂)₃]?
Conventionally, first complex ion is positively charged and the latter is negatively charged. Suppose thess charges are $p+$ and $q-$, respectively. Since two ion ratio is 1:1, $|p+|= |q-|$.

Suppose oxidation number of $\ce{Co}$ is $n$ and that of $\ce{Cr}$ is $m$. Hence, for cation complex (Note that the net charges of $\ce{NH3}$ and $\ce{NO2-}$ is $0$ and $-1$, respectively):
  $$n+0\times4+(-1)\times2=+p = n-2$$
and for anion complex:
 $$m+0\times3+(-1)\times3=-q=m-3$$
Since, $|p+|= |q-|$,
$$m-3= +p \; \text{and  }  n-2=-p$$
Or, $m+n=5$. 
The only answer agree with this is (A) or (B):

If (A) is correct, then, $\ce{Co^2+}$ and $\ce{Cr^3+}$. Substitute these values in original two equation: $+2+0\times4+(-1)\times2=+0 $ and $+3+0\times4+(-1)\times3=+0 $. This can't be.

If (B) is correct, then, $\ce{Co^3+}$ and $\ce{Cr^2+}$. Substitute these values in original two equation: $+3+0\times4+(-1)\times2=+1 $ and $+2+0\times4+(-1)\times3=-1 $. This is correct.

Therefore, your answer is (B), and the oxidation numbers of $\ce{Co}$ and $\ce{Cr}$ ared $+3$ and $+2$, respectively.