For a spring $kx=mg$
$x=\large\frac{mg}{k}$
$\quad= \large\frac{GMm}{k r^2}\qquad$$ \bigg[g= \large\frac{GM}{r^2}\bigg]$
$x\; \alpha \; \large\frac{1}{r^2}$
$\large\frac{x_2}{x_1}=\frac{R^2}{(R+h)^2}$
$x^2 =1 \times \bigg(\large\frac{6400}{7200}\bigg)^2$
$\quad= 0.79 \;cm$
Hence a is the correct answer.