Answer
$\lambda=523nm$
Work Step by Step
We know that;
$d sin({\theta})=m\lambda$
This can be written as:
$\lambda=\frac{d sin({\theta})}{m}$
We plug in the known values to obtain:
$\lambda=\frac{1.73\times 10^{-6}sin(17.6)}{1}=523\times 10^{-9}=523nm$
