Using Hooke's law for spring $F=-kx$

When the mass of $0.6 \mathrm{kg}$is suspended to the spring then spring force will balance the weight of suspended mass to the spring therefore $F=mg$ $\Rightarrow k=\frac{F}{x}=\frac{mg}{x}=\frac{0.6\times 10}{0.4}=15 \mathrm{N} {\mathrm{m}}^{-1}$

Time period for the period of oscillation of the string loaded by $255 \mathrm{gm} \text{or} 0.255 \mathrm{kg}$ is, $T=2\mathrm{\pi }\sqrt{\frac{\mathit{m}}{\mathit{k}}}=2\mathrm{\pi }\sqrt{\frac{0.255}{15}}=0.82 \mathrm{s}$