1. Interpret the two arrangements

For a mass–spring system, the time period is $T = 2\pi \sqrt{\frac{m}{k_{\text{eq}}}}$ where $k_{\text{eq}}$ is the effective spring constant.

2. Figure (a): springs in series

The springs have constants $k$ and $2k$.

For springs in series, $\frac{1}{k_{\text{eq,a}}} = \frac{1}{k} + \frac{1}{2k}$ $\frac{1}{k_{\text{eq,a}}} = \frac{2+1}{2k} = \frac{3}{2k}$ So, $k_{\text{eq,a}} = \frac{2k}{3}$

Given $T_a = 3\,\text{s}$, $3 = 2\pi \sqrt{\frac{m}{2k/3}} = 2\pi \sqrt{\frac{3m}{2k}}$

3. Figure (b): springs in parallel

For springs in parallel, $k_{\text{eq,b}} = k + 2k = 3k$

Hence, $T_b = 2\pi \sqrt{\frac{m}{3k}}$

4. Relate $T_b$ and $T_a$

Using ratio, $\frac{T_b}{T_a} = \sqrt{\frac{k_{\text{eq,a}}}{k_{\text{eq,b}}}}$

Substitute values: $\frac{T_b}{3} = \sqrt{\frac{(2k/3)}{3k}} = \sqrt{\frac{2}{9}} = \frac{\sqrt{2}}{3}$

Therefore, $T_b = 3 \cdot \frac{\sqrt{2}}{3} = \sqrt{2}\,\text{s}$

So if $T_b = \sqrt{x}\,\text{s}$, then $\sqrt{x} = \sqrt{2} \implies x=2$

5. Final answer

$\boxed{2}$

The derived answer matches the stored correct answer.