The temp of the furnance is 2324 degree celcius and intensity is maximum in the radiation spectrum nearly at 12000 angstrom.If the intensity in the spectrum of a star is maximum nearly at 4800 angstrom then surface temp of a star is.

There is an inverse relationship between the wavelength of the peak of the emission of a black body and its temperature, and this less powerful consequence is often also called Wien's displacement law

We can consider the star and the furnace both as blackbodies for this question.

(λmax)(T) = b in both cases.

Theresfore, (λmax)(T) for the furnace = (λmax)(T) for the star

Then, for the furnace, λmax = 12000 A = 1.2e-6 m and T =2324 C = 2597 K

For the star, (λmax) = 4800 A = 4.8e-7 m

Hence , Tstar = (1.2e-6)(2597)/(4.8e-7) = 6492.5 K