We know that $\mathrm{K}_{\max }$ (maximum energy) $=\mathrm{hv}-\phi$
As given that,
For the first case
Wavelength of light $\lambda_1=600 \mathrm{~nm}$
For the second case
Wavelength of light $\lambda_2=400 \mathrm{~nm}$
Let the maximum energies of emitted electron are $\mathrm{K}_1$ and $\mathrm{K}_2$
Also maximum kinetic energy for the second condition is equal to the twice of the kinetic energy in first condition.
i.e., ${ }_{\max 2}=2 \mathrm{~K} \max _1$
$$
\begin{array}{ll}
& \mathrm{K}_{\max _1}=\frac{\mathrm{hc}}{\lambda_1}-\phi \quad\left(\because \mathrm{K}_{\max }=\mathrm{hv}-\phi\right) \\
& \text { Here, } \mathrm{K}_{\max _2}=\frac{\mathrm{hc}}{\lambda_2}-\phi \\
\Rightarrow \quad & 2 \mathrm{~K}_{\max _1}=2\left(\frac{\mathrm{hc}}{\lambda_1}-\phi\right) \\
\text { Put } & \left(\mathrm{K}_{\max 1}\right) \text { in above equation } \\
\Rightarrow & 2\left(\frac{1230}{600}-\phi\right)=\left(\frac{1230}{400}-\phi\right) \quad[\because \mathrm{hc} \simeq 1230 \mathrm{eVnm}] \\
\Rightarrow & \phi=\frac{1230}{1200}=1.03 \mathrm{eV}
\end{array}
$$