Answer
0.33 volts.
Work Step by Step
The stopping potential $V_0$ is the voltage such that $eV_0$ equals the maximum kinetic energy.
Apply equation 27–5b to first find the work function.
$$KE_{max}=eV_0=\frac{hc}{\lambda_0}-W_0$$
$$ W_0= \frac{hc}{\lambda_0}- eV_0 $$
Now find the stopping potential for the higher wavelength light.
$$ eV_1=\frac{hc}{\lambda_1}-W_0= hc (\frac{1}{\lambda_1}-\frac{1}{\lambda_0})+eV_0$$
$$ eV_1= (1240 eV \cdot nm) (\frac{1}{440nm}-\frac{1}{270nm })+2.10eV=0.33eV$$
The new stopping potential is 0.33 volts.