To determine the optimum thickness of insulation for maximum heat dissipation from a current carrying wire, we need to consider the heat transfer mechanisms involved.

Firstly, we have to calculate the heat dissipation through convection from the wire to the surrounding air. This can be done using the formula:

Q_conv = h * A * ΔT

Where:
Q_conv is the heat dissipation through convection
h is the convective heat transfer coefficient (20 W/m2K for air)
A is the surface area of the wire exposed to air
ΔT is the temperature difference between the wire surface and the surrounding air

The surface area of the wire can be calculated using the formula:

A = π * D * L

Where:
D is the diameter of the wire (20 mm)
L is the length of the wire

Next, we need to calculate the heat dissipation through conduction from the wire to the insulation. This can be done using the formula:

Q_cond = k * A * ΔT / δ

Where:
Q_cond is the heat dissipation through conduction
k is the thermal conductivity of the insulation material (0.5 W/mK)
A is the surface area of the wire exposed to the insulation
ΔT is the temperature difference between the wire surface and the insulation
δ is the thickness of the insulation

To maximize heat dissipation, we need to minimize the resistance to heat transfer. The resistance to heat transfer through convection is inversely proportional to the convective heat transfer coefficient, while the resistance to heat transfer through conduction is directly proportional to the thermal conductivity and the thickness of the insulation.

Therefore, for maximum heat dissipation, we need to minimize the resistance to heat transfer through conduction. This means that we need to minimize the thickness of the insulation. The thinner the insulation, the lower the resistance to heat transfer and the higher the heat dissipation.

In this case, the correct answer is option 'D' (15 mm) because it represents the thinnest insulation option among the given choices.