b
\(\text { Given } \mathrm{R}_0=8 \Omega, \mathrm{R}_{100}=10 \Omega\)

\(\therefore \mathrm{R}_{100}=\mathrm{R}_0(1+\alpha \Delta \mathrm{T})\)

\(\text { Also, } \mathrm{R}_{400}=\mathrm{R}_0\left(1+\alpha \Delta \mathrm{T}^1\right)\)

\(\therefore 10=8(1+\alpha \times 100) \Rightarrow 100 \alpha=\frac{1}{4}\)

\(\therefore \mathrm{R}_{400}=8(1+400 \alpha)=8(1+1)=16 \Omega\)

Hence option \((2)\)