Answer
9666.67 $ \times 10^{-9}$ m or 9666.67 nm
Work Step by Step
Temperature of the walls of blackbody cavity = 27 $^{0}C$
absolute kelvin temperature = 273 + temperature in $^{0}C$
so temperature of blackbody cavity in kelvin = 273 + 27 $^{0}C$ = 300 $^{0}K$
from Wien's displacement law $\lambda_{max} T = 2.90\times 10^{-3} $ m.K
so $\lambda_{max} = \frac{2.90\times 10^{-3} m.K} {T}$
putting the value of blackbody temperature $\lambda_{max} = \frac{2.90\times 10^{-3} m.K} {300 K}$
$\lambda_{max}= 9666.67\times 10^{-9}$ m
$\lambda_{max} =9666.67 nm
