You have mentioned the right equations. But you have to think carefully about the phrase" they have a point of superposition at $\displaystyle 20^o$". For the two light rays of different wavelengths to superimpose with each other, they must first meet each other.
$\displaystyle \frac{n_1 \lambda_1}{d}=\frac{n_2 \lambda_2}{d}$
$\displaystyle n_1 \lambda_1=n_2 \lambda_2$
For n to be an interger, in fact, you are finding the L.C.M. of 400nm and 600nm which is 1200nm
When you plug $\displaystyle n_1=3,\lambda_1=400nm, d=1/285000m^{1}$, you find $\displaystyle \theta=20^o$
When you plug $\displaystyle n_2=2,\lambda_1=600nm, d=1/285000m^{1}$, you find $\displaystyle \theta=20^o$
So, from these, the second time they superimpose with each other, it should be 2400nm, for the following work, you can handle it.
