Answer
$\lambda_{max} = 9.35\times 10^{-6}~m$
Work Step by Step
We can convert the temperature to units of Kelvin:
$T = \frac{5}{9}(F-32)+273$
$T = \frac{5}{9}(98.6-32)+273$
$T = 310~K$
We can find the wavelength at which the spectral radiancy is maximum:
$\lambda_{max}~T = 2.898\times 10^{-3}~m~K$
$\lambda_{max} = \frac{2.898\times 10^{-3}~m~K}{T}$
$\lambda_{max} = \frac{2.898\times 10^{-3}~m~K}{310~K}$
$\lambda_{max} = 9.35\times 10^{-6}~m$
