Answer
At $71^{\circ}C$ the resistance of the wire is $0.30 Ω$
Work Step by Step
$ρ = ρ_{^{\circ}}[1+ \alpha (T-T_{^{\circ}})]$
Substitute the values in the above equation with given values,
$0.30Ω = 0.25Ω [1 + 3.9 \times (10^{-3})^{\circ}C^{-1}(T - 20^{\circ}C)]$
Solve for T
$T = 71.28^{\circ}C \approx 71^{\circ}C$
