No. Holding constant the electrical power heating the source is very different than holding constant the source temperature. Like Jane, let's assume the source temperature is constant (rather than the electrical heating power) and use Jane's equation and notation:
... we have 4 surfaces, which I will call 1, 2, 3, 4 moving outward, so 1 is the surface of the heat source, 2 the inside of the hollow sphere, 3 the outside of the hollow sphere, and 4 the chamber wall. T3 for example would be radiative Temperature of surface 3. ... [Jane Q. Public, 2014-09-10]
Draw a boundary between the source (T1=150F) and the chamber walls (T4=0F) before the hollow sphere is added. Power in = power out. Variable "electricity_initial" flows in at whatever rate is needed to keep T1=150F. Net heat transfer flows out from source to chamber walls. Power in = power out:
electricity_initial = p(14) = (e)(s) * ( T1^4 - T4^4 ) = (e)(s) * (8908858139.78) = 55.5913 W/m^2
Now add the hollow sphere and draw a boundary between the source (T1=150F) and the inside of the hollow sphere (T2). A different "electricity_final" flows in, and heat transfer p(12) flows out.
electricity_final = p(12) = (e)(s) * ( T1^4 - T2^4 )
Now draw a boundary between the outside of the hollow sphere (T3=T2) and the chamber walls (T4=0F): "electricity_final" flows in, and heat transfer p(34) flows out. Since power in = power out:
electricity_final = p(34) = (e)(s) * ( T2^4 - T4^4 )
Combine these two equations:
T1^4 - T2^4 = T2^4 - T4^4
T2 = T3 = 305.47K = 90.176 deg. F.
electricity_final = 27.8 W/m^2.
So if the source temperature is held constant at 150F, adding the hollow sphere reduces the necessary electrical heating power to keep the source at 150F by a factor of two, from 55.6 to 27.8 W/m^2.
Can we agree on that?