The potential difference that must be applied to stop the fastest...

Energy of incident light $\mathrm{E}(\mathrm{eV})=\frac{12375}{2000}$ $=6.2 \mathrm{eV}(200 \mathrm{~nm}=2000 Å)$ According to the relation $\mathrm{E}=\mathrm{W}_0+\mathrm{eV}_0$ \begin{aligned} \Rightarrow \quad \mathrm{V}_0 & =\frac{\mathrm{E}-\mathrm{W}_0}{\mathrm{e}} \\ & =\frac{(6.2-5.01) \mathrm{e}}{\mathrm{e}} \\ & =1.2 \mathrm{~V} \end{aligned}