The work functions for metals A, B and C are respectively 1.92 eV,...

That metal will emit photoelectrons which has work function lower than that obtained with the radiation of $4100 Å$. Work function for wavelength of $4100 Å$ is $$\begin{aligned} W & =\frac{h c}{\lambda} \\ & =\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{4100 \times 10^{-10}} \\ & =4.8 \times 10^{-19} \mathrm{~J} \\ & =\frac{48 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}=3 \mathrm{eV}\end{aligned}$$ Now, we have $$\begin{aligned}& W_A=1.92 \mathrm{eV} \& W_B=2.0 \mathrm{eV} \\ & W_C=5 \mathrm{eV}\end{aligned}$$ Since, $W_A \lt W$ and $W_B \lt W$, hence, $A and B$ will emit photoelectrons.