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Explanation:

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a) I₂ = 2 mA   (The current has decreased)

b) L₂ = 2.4 cm

Explanation:

Consider the old wire as wire 1 and the new wire as wire 2. The data that we have from the question is:

Current through wire 1 = I₁ = 12.5 mA

Diameter of wire 1 = d₁ = 5 mm

Length of wire 1 = L₁ = 15 cm

Cross-Sectional Area of wire 1 = A₁ = πd₁²/4 = π(5 mm)²/4 = 19.63 mm²

Diameter of wire 2 = d₂ = 2 mm

Cross-Sectional Area of wire 2 = A₂ = πd₂²/4 = π(2 mm)²/4 = 3.14 mm²

a)

Length of wire 2 = L₂ = 15 cm

Since, the battery is same. Therefore, the voltage will be same for both wires.

V₁ = V₂

using Ohm's Law (V = IR)

I₁R₁ = I₂R₂

Since resistance of wire is given by formula:  R = ρL/A

Therefore,

I₁ρ₁L₁/A₁ = I₂ρ₂L₂/A₂

where,

ρ = resistivity, and it depends upon material of wire and the material of both wires is same and the length of wires is also same.

Hence,    ρ₁ = ρ₂

and      L₁ = L₂

and the equation becomes:

I₁/A₁ = I₂/A₂

I₂ = I₁A₂/A₁

I₂ = (12.5 mA)(3.14 mm²)/(19.63 mm²)

I₂ = 2 mA

Thus, the current has decreased.

b)

In order to have same current the resistance of both wires must be same:

R₁ = R₂

ρ₁L₁/A₁ = ρ₂L₂/A₂

Since,   ρ₁ = ρ₂

Therefore,

L₁/A₁ = L₂/A₂

L₂ = L₁A₂/A₁

L₂ = (15 cm)(3.14 mm²)/(19.63 mm²)

L₂ = 2.4 cm