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

Initially, the period is $1\,\text{s}$: $2\pi\sqrt{\frac{m}{k}}=1 \quad \text{(1)}$

When the mass is increased by $3\,\text{kg}$, the new period increases by $1\,\text{s}$. So the new period is $T'=2\,\text{s}$ Hence, $2\pi\sqrt{\frac{m+3}{k}}=2 \quad \text{(2)}$

Divide equation (2) by equation (1): $\frac{2\pi\sqrt{\frac{m+3}{k}}}{2\pi\sqrt{\frac{m}{k}}}=\frac{2}{1}$ $\sqrt{\frac{m+3}{m}}=2$

Squaring both sides, $\frac{m+3}{m}=4$ $m+3=4m$ $3=3m$ $m=1\,\text{kg}$

Therefore, the required mass is $\boxed{1}$ kg.