In a single slit diffraction experiment first minimum for red light (660 nm) coincides with first maximum of some other wavelength l'. The value of l' is 

In a single slit diffraction experiment, position of minima is given by  $d\sin \theta =n\lambda$
So for first minima of red   $\sin \theta =1\times \left(\frac{{\lambda }_R}{d}\right)$
and as first maxima is midway between first and second minima, for wavelength ${\lambda }^{\text{'}}$,
its position will be 
$\Rightarrow \sin {\theta }^{\text{'}}=\frac{3{\lambda }^{\text{'}}}{2d}$
According to given condition  $\sin \theta =\sin {\theta }^{\text{'}}$
 $\Rightarrow {\lambda }^{\text{'}}=\frac{2}{3}{\lambda }_R$

So ${\lambda }^{\text{'}}=\frac{2}{3}\times 660=440nm$$=4400\AA$