Part (a): head in feet of oil
$h_{oil} = \dfrac{p}{\gamma_{oil}}$
$h_{oil} = \dfrac{p}{s_{oil} \, \gamma_{water}}$
$h_{oil} = \dfrac{60(12^2)}{0.80(62.4)}$
$h_{oil} = 173.08 \, \text{ft}$ answer
Part (b) head in feet of water
$h_{water} = s_{oil} \, h_{oil}$
$h_{water} = 0.80(173.08)$
$h_{water} = 138.46 \, \text{ft}$ answer
Part (c) head in inches of mercury
$h_{mercury} = \dfrac{s_{oil}}{s_{mercury}}h_{oil}$
$h_{mercury} = \dfrac{0.80}{13.6}(173.08)$
$h_{mercury} = 10.18 \, \text{ft}$
$h_{mercury} = 122.17 \, \text{in}$ answer