As per given figures, two springs of spring constants K and 2K are...

In figure (a), the springs are connected in series. Therefore,

${K}_{eq} = \dfrac{{K}_{1}{K}_{2}}{{K}_{1} + {K}_{2}} = \dfrac{K \times 2K}{K + 2K} = \dfrac{2K}{3}$

Therefore, the time period

${T}_{a} = 2\pi\sqrt{\dfrac{m}{{K}_{eq}}} = 2\pi\sqrt{\dfrac{m}{\dfrac{2K}{3}}} = 2\pi\sqrt{\dfrac{3m}{2K}}$

In figure (b), the springs are connected in parallel. Therefore,

${K}_{eq} = {K}_{1} + {K}_{2} = 3K$

Therefore, the time period

${T}_{b} = 2\pi\sqrt{\dfrac{m}{{K}_{eq}}} = 2\pi\sqrt{\dfrac{m}{3K}}$

Now,

$\dfrac{{T}_{a}}{{T}_{b}} = \sqrt{\dfrac{9}{2}} ⇒ {T}_{b} = \dfrac{\sqrt{2}}{3}{T}_{a} = \sqrt{2}s$

Hence, $x = 2$