Correct Answer - B
Potential difference applied `= V = IR`
`= 30 xx 10^(-3)xx10`
`=0.3V`
Resistance at `120^(@)C` is given by,
`R_(120)=R_(20)(1+alpha xx 100)=15 Omega`.
`:.` Current at `120^(@)C` is given by,
`I=(V)/(R_(120))=(0.3)/(15)=0.02 A = 20 mA`