Answer
$667.8\;\rm nm$
Work Step by Step
Since we have several wavelengths, each wavelength will diffract at a different angle.
We are given that the light of 501.5 nm creates a first-order bright fringe 21.90 cm from the central maximum.
Hence, for this wavelength,
$$d\sin\theta_1=(1)\lambda$$
Now we need to solve for $d$,
$$d=\dfrac{\lambda}{\sin\theta_1}$$
where $\sin\theta_1=\sin(\tan^{-1}\theta_1)=\sin(\tan^{-1}[y_1/L])$
$$d=\dfrac{\lambda }{\sin(\tan^{-1}[y_1/L])}\tag 1$$
For the other wavelength,
$$d\sin\theta_1'=(1)\lambda'$$
Now we need to solve for $\lambda'$
$$ \lambda'=d\sin\theta_1'$$
where $\sin\theta_1'=\sin(\tan^{-1}[y_1'/L])$
$$ \lambda'=d \sin(\tan^{-1}[y_1'/L])$$
Plugging from (1);
$$ \lambda'=\dfrac{\lambda }{\sin(\tan^{-1}[y_1/L])}\sin(\tan^{-1}[y_1'/L])$$
Plugging the given;
$$ \lambda'=\dfrac{501.5}{\sin(\tan^{-1}[21.90/50.0])}\sin(\tan^{-1}[31.60/50.0])$$
$$ \lambda'=\color{red}{\bf 667.8}\;\rm nm$$
