1. Use Einstein’s photoelectric equation

For a metal with threshold frequency $\nu_0$, the maximum kinetic energy of emitted electrons is

$K_{\max}=h\nu-h\nu_0=h(\nu-\nu_0).$

Also,

$K_{\max}=\frac{1}{2}mv^2.$

So,

$\frac{1}{2}mv^2=h(\nu-\nu_0).$

2. Case 1: incident frequency is twice the threshold frequency

Given

$\nu=2\nu_0.$

Then

$\frac{1}{2}mv_1^2=h(2\nu_0-\nu_0)=h\nu_0.$

So,

$v_1^2=\frac{2h\nu_0}{m}.$

3. Case 2: incident frequency is five times the threshold frequency

Given

$\nu=5\nu_0.$

Then

$\frac{1}{2}mv_2^2=h(5\nu_0-\nu_0)=4h\nu_0.$

So,

$v_2^2=\frac{8h\nu_0}{m}.$

4. Find the ratio $x$ where $v_2=xv_1$

$\left(\frac{v_2}{v_1}\right)^2=\frac{v_2^2}{v_1^2}=\frac{\frac{8h\nu_0}{m}}{\frac{2h\nu_0}{m}}=4.$

Hence,

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

Therefore,

$x=2.$

5. Comparison with stored answer

Stored correct answer: $2$

Derived answer: $2$

They match.