If the frequency of incident light is tripled, the stopping potential will

The relation between stopping potential (V₀) and frequency of incident ($\nu$) radiation is given by

${\mathrm{V}}_0 = \frac{\mathrm{h}\nu }{\mathrm{e}} - \frac{\mathrm{\phi }}{\mathrm{e}}$

When the frequency is tripled, 

$$\begin{gathered} {\mathrm{V}}_0\text{'} = \frac{3\mathrm{h}\nu }{\mathrm{e}} - \frac{\mathrm{\phi }}{\mathrm{e}} \\ \Rightarrow {\mathrm{V}}_0\text{'} = \frac{3\mathrm{h}\nu }{\mathrm{e}} - \frac{3\mathrm{\phi }}{\mathrm{e}} + \frac{3\mathrm{\phi }}{\mathrm{e}} - \frac{\mathrm{\phi }}{\mathrm{e}} \\ \Rightarrow {\mathrm{V}}_0\text{'}= 3\left(\frac{\mathrm{h}\nu }{\mathrm{e}} - \frac{\mathrm{\phi }}{\mathrm{e}}\right) + \frac{2\mathrm{\phi }}{\mathrm{e}} \\ \Rightarrow {\mathrm{V}}_0\text{'} = 3{\mathrm{V}}_0+ \frac{2\mathrm{\phi }}{\mathrm{e}} \end{gathered}$$

Hence, if the frequency of intensity of light is tripled, the stopping potential will become more than tripled.