Einstein's photoelectric equation

When light photons having energy $E=2.96\text{eV}$‍ strike a ceasium surface, the most energetic electrons have kinetic energy $0.80\text{eV}$‍.

Find the minimum frequency of light for which photoelectric emission takes place from the caesium surface.$\begin{array}{rl}& \\ & \end{array}$‍ Note: Take $h=6.63\times {10}^{-34}\text{J s}$‍ and $e=1.60\times {10}^{-19}\text{C}$‍ and round-off your answer to three significant figures.

$\times {10}^{14}\text{Hz}$‍