\(a\sin \theta  = \left( {2n + 1\frac{\lambda }{2}} \right)\)

and for diffraction minima,

\(a \sin \theta=n \lambda\)

According to question,

\(\left(2 \times 1+1 \frac{\lambda}{2}\right)=1 \times 6600\)

\(\left( {\because {\lambda _{\text{R}}} = 6600\,\,\mathop {\text{A}}\limits^o } \right)\)

\({\lambda  = \frac{{6600 \times 2}}{3}}\)

\(\,\lambda  = 4400\,\mathop {\text{A}}\limits^o \)