Answer
$1581\;\rm nm$
Work Step by Step
We know that the intensity of scattered light $I$ is inversely proportional to ${\lambda^4}$.
So
$$I\propto \dfrac{1}{\lambda^4}$$
Hence,
$$I_{\rm 500nm}\propto \dfrac{1}{500^4}$$
$$I_{\rm IR}\propto\dfrac{1}{\lambda_{IR}^4}$$
Divide $I_{\rm 500nm}$ by $I_{IR}$;
$$\dfrac{I_{\rm 500nm}}{I_{IR}}=\dfrac{\dfrac{1}{500^4}}{\dfrac{1}{\lambda_{IR}^4}}$$
and we are given that $I_{\rm IR}=0.01I_{\rm 500nm}$
$$\dfrac{ \color{red}{\bf\not} I_{\rm 500nm}}{0.01 \color{red}{\bf\not} I_{\rm 500nm}}= \dfrac{\lambda_{IR}^4}{500^4} $$
$$100= \dfrac{\lambda_{IR}^4}{500^4} $$
$$500^4\times 100= \lambda_{IR}^4 $$
Thus,
$$ \lambda_{IR} =\sqrt[4]{500^4\times 100} =\color{red}{\bf 1581}\;\rm nm$$
