Answer
$432.3nm$
Work Step by Step
We know that
$\nu=\frac{E}{h}=\frac{4.60\times 10^{-19}}{6.63\times 10^{-34}}=6.938\times 10^{14}s^{-1}$
We also know that
$\lambda=\frac{c}{\nu}$
$\lambda=\frac{3.00\times 10^8}{6.93\times 10^{14}}=4.323\times 10^{-7}m=432.3nm$