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

$K_{\mathrm{eq}}=\frac{K_1{K}_2}{K_1+K_2}=\frac{K\times 2K}{K+2K}=\frac{2K}{3}$

Therefore, the time period

$T_a=2\pi \sqrt{\frac{m}{K_{\mathrm{eq}}}}=2\pi \sqrt{\frac{m}{{\displaystyle \frac{2K}{3}}}}=2\pi \sqrt{\frac{3m}{2K}}$

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

$K_{eq}=K_1+K_2=3K$

Therefore, the time period

$T_b=2\pi \sqrt{\frac{m}{K_{\mathrm{eq}}}}=2\pi \sqrt{\frac{m}{3K}}$

Now,

$\frac{T_a}{T_b}=\sqrt{\frac{9}{2}}\Rightarrow T_b=\frac{\sqrt{2}}{3}{T}_a=\sqrt{2} \mathrm{s}$

Hence, $x=2$