d
$i_0 R_0=i_{100} R_{100} \quad$ [For same source]
$\Rightarrow 2 R_0=1.2 R_0[1+100 \alpha]$
$\Rightarrow 1+100 \alpha=\frac{5}{3} \Rightarrow 100 \alpha=\frac{2}{3}$
$\Rightarrow 50 \alpha=\frac{1}{3}$
$\therefore i_{50} R_{50}=i_0 R_0$
$\Rightarrow i_{50}=\frac{i_0 R_0}{R_{50}}=\frac{2 \times R_0}{R_0(1+50 \alpha)}=\frac{2}{1+\frac{1}{3}}=1.5\,A$
$=15 \times 10^2\,mA$