let $X$ be some random variable with density $f$.
The mean of $X$ is $E(X)=\displaystyle\int_{-\infty}^{\infty}xf(x)dx=\int_{1}^{\infty}2x^{-2}dx=(-\frac2x|_{1}^{\infty}=2$.
And the median of $X$ is $M$ such that $\displaystyle\int_{-\infty}^{M}f(x)dx = \frac12$.
$\displaystyle\int_{-\infty}^{M}f(x)dx = \int_{1}^{M}2x^{-3}dx = (-x^{-2}|_{1}^{M} = 1-\frac1{M^2}$
therefore $\displaystyle\frac1{M^2}=\frac12$ which means $M=\sqrt2$