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Assume emissivity and absorptivity of the heated material are equal. Assume a perfectly insulated material with no conduction or convection, so all heat transfer will occur as radiation.

Stefan-Boltzmann constant (sigma): 5.67 x 10^-8 W/(m^2*K^4)
Area of the material's surface exposed to radiate: 4 in^2 = 0.00258 m^2
Area of the lens = 1200 in^2 = 0.7742 m^2
Incident radiated heat energy transfer rate (Q_dot_incident): 1000 W/m^2 * area of lens = 774 W (no losses)
T_s = absolute temperature of the surface (in Kelvin)

The material will radiate heat with rate:
Q_dot_emit = emissivity * sigma * area of material * T_s^4

And absorb heat with rate:
Q_dot_absorb = absorptivity * Q_dot_incident

Equating the two to solve for a final, steady state temperature:
T_s = (Q_dot_incident/(sigma * area of material))^.25
= (774/(5.67E-8 * 0.00258))^.25
= 1517 K = 1244 ºC = 2270 ºF