Absolute Pressure at 200 mm Below the Surface of Liquid Mercury

$p_{absolute} = p_{atmospheric} + p_{gauge}$

$p_{absolute} = 101.325 + (\gamma h)_\text{mercury}$

$p_{absolute} = 101.325 + s_\text{mercury}\gamma_\text{water}h_\text{mercury}$

$p_{absolute} = 101.325 + 13.6(9.81)(0.2)$

$p_{absolute} = 128.0082 ~ \text{kPa}$   ←   Answer: [ C ]