Critical radius of insulation refers to the thickness of insulation at which the heat transfer through the insulation is minimized. It is the point at which the heat transfer by conduction through the insulation material is equal to the heat transfer by convection from the outer surface of the insulation to the surrounding air.

Given data:
Inner diameter of the steel steam pipe (Di) = 10 cm
Outer diameter of the steel steam pipe (Do) = 11 cm
Thermal conductivity of insulation (k) = 1 W/mK
Convective heat transfer coefficient (h) = 8 W/m2K

To determine the critical radius of insulation, we need to equate the heat transfer by conduction through the insulation to the heat transfer by convection from the outer surface of the insulation.

1. Heat transfer by conduction through the insulation:
The heat transfer rate by conduction through the insulation can be calculated using the formula:
Qcond = (k * A * ΔT) / L

Where:
Qcond = Heat transfer rate by conduction (W)
k = Thermal conductivity of insulation (W/mK)
A = Surface area of the insulation (m2)
ΔT = Temperature difference between the inner and outer surfaces of the insulation (K)
L = Thickness of the insulation (m)

2. Heat transfer by convection from the outer surface of the insulation:
The heat transfer rate by convection can be calculated using the formula:
Qconv = h * A * ΔT

Where:
Qconv = Heat transfer rate by convection (W)
h = Convective heat transfer coefficient (W/m2K)
A = Surface area of the insulation (m2)
ΔT = Temperature difference between the outer surface of the insulation and the surrounding air (K)

To determine the critical radius of insulation, we need to find the thickness of insulation (L) at which Qcond = Qconv.

Let's assume the critical radius of insulation as 'r' and the thickness of insulation as 'L'. Therefore, the inner radius of the insulation will be (r - L) and the outer radius of the insulation will be 'r'.

Surface area of the insulation (A) can be calculated as:
A = 2π * r * L

Temperature difference (ΔT) can be assumed as the temperature difference between the inner surface of the insulation and the surrounding air.

Now, equating the heat transfer by conduction and convection:
(k * A * ΔT) / L = h * A * ΔT

Canceling out common terms:
k / L = h

Substituting the known values:
1 / L = 8

Solving for L:
L = 1 / 8 = 0.125 m = 12.5 cm

Therefore, the critical radius of insulation is 12.5 cm, which corresponds to option 'C'.