Answer
$4.84\times10^{14}\,s^{-1}$
Work Step by Step
Binding energy of the emitted electron $\phi=\frac{193\times10^{3}\,J}{6.022\times10^{23}}=3.2049\times10^{-19}\,J$
$\phi=h\nu\implies \text{Threshold frequency }\nu=\frac{\phi}{h}$
$=\frac{3.2049\times 10^{-19}\,J}{6.626\times10^{-34}\,J\cdot s}$
$=4.84\times10^{14}\,s^{-1}$
