The relation between the stopping potential and the frequency of the incident light is given by

$e{V}_0=h\nu -\varphi ...\left(1\right)$

When $V_{0 }= 0$,

$\varphi =h\nu ...\left(2\right)$

Substitute the value of the frequency corresponding to zero stopping potential into equation (2) to calculate the required work function of the material.

$\begin{array}{rcl}\mathrm{\varphi }& =& \frac{6.626\times {10}^{-34} \mathrm{J}·\mathrm{s}\times 5\times {10}^{14} \mathrm{Hz}}{1.6\times {10}^{-19} \mathrm{J}/\mathrm{eV}}\\ & =& \frac{6.626\times 5}{16}\approx 2.07\mathrm{eV}\end{array}$