When 5V potential difference is applied across a wire length 0.1 m, the drift speed of electron is 2.5 × 10⁻⁸ ms⁻¹. If the electron density in the wire is 8 × 10²⁸ m⁻³, the resistivity of the material is close to 1.6 × 10⁻⁸ Ωm, 1.6 × 10⁻⁷ Ωm, 1.6 × 10⁻⁹ Ωm, 1.6 × 10⁻⁶ Ωm

Step 1: Current flowing through the wire
We know that
I = nAevd
We also know by ohm's law , I = V/R ⇒ V/R = Anevd (1)
Step 2: Resistance of the wire
R = ρl/A (2)
Step 3: Solving above equations
From (1) and (2)
V(ρl/A) = Anevd ⇒ VAρl = Anevd ⇒ ρ = Vl/nevd
= 5V × 0.1m / (8 × 10²⁸m⁻³ × 1.6 × 10⁻¹⁹C × 2.5 × 10⁻⁸ms⁻¹)
= 1.5625 × 10⁻⁸ Ωm
⇒ ρ ≈ 1.6 × 10⁻⁸ Ωm
Hence, option (C) is correct.