By Ohm's law, V=IR.
Let the current and resistance at 0∘C be {i}_{0}&{R}_{0} respectively.
Let the current and resistance at 100∘C be {i}_{100}&{R}_{100} respectively.
By the given condition,
i0R0=i100R100⇒2R0=1.2R0(1+100α)⇒1+100α=35⇒50α=31
Since the voltage remains constant,
i50R50=i0R0⇒i50=R50i0R0=R0(1+50α)2×R0=1+312=1.5A⇒i50=15×102mA