Answer
$\phi = 7.3\times 10^{-19}~J$
Work Step by Step
We can find the magnitude of the energy removed from the electron by the stopping potential:
$E = q~\Delta V$
$E = (1.6\times 10^{-19}~C)(1.1~V)$
$E = 1.76\times 10^{-19}~J$
This energy is equal to the maximum kinetic energy of the electron as it is ejected from the tungsten surface. We can find the work function $\phi$:
$K_{max} = \frac{hc}{\lambda}-\phi$
$\phi = \frac{hc}{\lambda}-K_{max}$
$\phi = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~/s)}{220\times 10^{-9}~m}-1.76\times 10^{-19}~J$
$\phi = 9.035\times 10^{-19}~J-1.76\times 10^{-19}~J$
$\phi = 7.3\times 10^{-19}~J$
