Answer
$3.68mm$
Work Step by Step
We can find the required how far the mercury rises from its initial position as follows:
The difference in height is given as
$dh=\frac{\rho_w h_w}{2\rho_{Hg}}$
We plug in the known values to obtian:
$dh=\frac{(1000)(0.1)}{2(13600)}$
$\implies dh=3.68mm$
