[tex]\lambda_\text{max} = 2.63\times 10^{-9}\;\text{m}[/tex].
The peak emission wavelength of an object depends on its absolute temperature.
[tex]\lambda_\text{max} = \dfrac{2.90\times 10^{-3}}{T}[/tex],
where
For the gas falling into the black hole,
[tex]T = 1.10\times 10^{6}\;\text{K}[/tex].
Apply the formula:
[tex]\lambda_\text{max} = \dfrac{2.90\times 10^{-3}}{T} = \dfrac{2.90\times 10^{-3}}{1.10\times 10^{6}} = 2.64\times 10^{-9}\;\text{m} = 2.64 \;\text{nm}[/tex].
The question mentioned that [tex]\lambda_\text{max}[/tex] is in the X-ray region of the electromagnetic spectrum. According to Encyclopedia Britannica, the wavelength of X-rays range from [tex]10^{-8}\;\text{m}=10\;\text{nm}[/tex] to [tex]10^{-10}\;\text{m} = 0.1\;\text{nm}[/tex], which indeed includes [tex]2.64\times 10^{-9}\;\text{m} = 2.64 \;\text{nm}[/tex].