Answer
$\lambda=469~nm$
Work Step by Step
We can find the required wavelength as follows:
$\lambda=\frac{dsin\theta}{m}$......eq(1)
But we know that
$tan\theta=\frac{y}{L}$
$\implies \theta=ta^{-1}(\frac{y}{L})$
We plug in the values of $\theta$ in eq(1) and $d=\frac{1}{N}$ to obtain:
$\lambda=\frac{sin(tan^{-1}(\frac{y}{L}))}{mN}$
We plug in the known values to obtain:
$\lambda=\frac{sin(tan^{-1}(\frac{0.164}{1}))}{1(345000)}$
$\lambda=469~nm$
