Answer
$2.13eV$, or $3.41\times 10^{-19}J$
Work Step by Step
First we need to find the work function of the material.
$1.1eV = \frac{hc}{400 nm} = \frac{1240 eV-nm}{400 nm}-\phi$
$1.1eV = 3.1 eV - \phi$
$\phi = 2eV$
Now we just use our new wavelength
$K = \frac{hc}{300 nm} - 2eV = \frac{1240 eV-nm}{300nm}-2eV$
$=2.13eV$
