Light of wavelength 330 nm falling on a piece of metal ejects electrons with sufficient energy which requires voltage $V_0$ to prevent them from reaching a collector. In the same setup, light of wavelength 220 nm, ejects electrons which require twice the voltage $V_0$ to stop them in reaching a collector. The numerical value of voltage $V_0$ is:

$\frac{15}{8}V$

Let W be the work function of metal. Then,

$eV_0=\frac{hc}{330×10^{-9}}-W$   (i)

$e(2V_0)=\frac{hc}{220×10^{-9}}-W$   (ii)

Solving these two equations, we get

$V_0=\frac{10^9×h×c}{110×e×6}=\frac{10^9×6.6×10^{-34}×3×10^8}{110×1.6×10^{-19}×6}=\frac{15}{8}V$