Answer
$P_{w}-P_{sw}=3.39kPa$
Work Step by Step
Assuming te effect of aire column on pressure is negligible and solving the manometer:
$P_{w}+\rho_{w}*g*h_{w}-\rho_{Hg}*g*h_{Hg}+\rho_{sw}*g*h_{sw}=P_{sw}$
$P_{w}-P_{sw}=\rho_{Hg}*g*h_{Hg}-\rho_{w}*g*h_{w}-\rho_{sw}*g*h_{sw}$
$P_{w}-P_{sw}=g*(\rho_{Hg}*h_{Hg}-\rho_{w}*h_{w}-\rho_{sw}*h_{sw})$
Substituting:
$P_{w}-P_{sw}=(9.81\frac{m}{s^2})*(13600\frac{kg}{m^3}*0.1m-1000\frac{kg}{m^3}*0.6-1035\frac{kg}{m^3}*0.4)$
$P_{w}-P_{sw}=3.39kPa$
