1. Use Einstein’s photoelectric equation

For incident photon energy $E$ and work function $\phi$,

$K_{\max}=E-\phi$

Also,

$K_{\max}=\frac{1}{2}mv_{\max}^2$

So the maximum speed satisfies

$v_{\max}\propto \sqrt{E-\phi}$

2. Write expressions for the two cases

For $E_1=4\,\text{eV}$,

$\frac{1}{2}mv_1^2=4-\phi$

For $E_2=2.5\,\text{eV}$,

$\frac{1}{2}mv_2^2=2.5-\phi$

Given,

$\frac{v_1}{v_2}=2$

Squaring both sides,

$\frac{v_1^2}{v_2^2}=4$

Since $v^2 \propto K_{\max}$,

$\frac{4-\phi}{2.5-\phi}=4$

3. Solve for $\phi$

$4-\phi=4(2.5-\phi)$

$4-\phi=10-4\phi$

$3\phi=6$

$\phi=2\,\text{eV}$

4. Final answer

The work function of the metal is

$\boxed{2\,\text{eV}}$