c
According to Einstein's photoelectric equation,

\(e V_{s}=\frac{h c}{\lambda}-\frac{h c}{\lambda_{0}}\)

\(\therefore\) As per question, \(e V=\frac{h c}{\lambda}-\frac{h c}{\lambda_{0}}\)   ..... \((i)\)

\(\frac{e V}{4}=\frac{h c}{2 \lambda}-\frac{h c}{\lambda_{0}}\)   ..... \((ii)\)

From equations \((i)\) and \((ii)\), we get

\(\frac{h c}{2 \lambda}-\frac{h c}{4 \lambda}=\frac{h c}{\lambda_{0}}-\frac{h c}{4 \lambda_{0}}\)

\(\Rightarrow \frac{h c}{4 \lambda}=\frac{3 h c}{4 \lambda_{0}}\) or \(\lambda_{0}=3 \lambda\)