Answer
$E_0 = 1.77~eV$
Work Step by Step
When $\lambda = 700~nm$, we can assume that $K_{max} = 0$
We can find the maximum work function of the cathode:
$K_{max} = \frac{h~c}{\lambda} - E_0$
$E_0 = \frac{h~c}{\lambda}$
$E_0 = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{700\times 10^{-9}~m}$
$E_0 = 2.84\times 10^{-19}~J$
$E_0 = (2.84\times 10^{-19}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$
$E_0 = 1.77~eV$