Answer
We can rank the barrels in order of the force on the plug due to the liquid in the barrel, from largest to smallest:
$c = e \gt b = d \gt a$
Work Step by Step
The pressure on the plug from the inside of the barrel is $P_{atm}+ P_{gauge}$ while the pressure on the plug from outside the barrel is $P_{atm}$. Therefore, the net pressure on the plug is $P_{gauge}$. We can calculate the force on the plug for each barrel.
(a) $F = P_g~A$
$F = [\rho~g~(h-0.20~m)](\pi~r^2)$
$F = [(1000~kg/m^3)~(9.80~m/s^2)~(1.00~m-0.20~m)](\pi)~(0.010~m)^2$
$F = 2.46~N$
(b) $F = P_g~A$
$F = [\rho~g~(h-0.20~m)](\pi~r^2)$
$F = [(1000~kg/m^3)~(9.80~m/s^2)~(1.20~m-0.20~m)](\pi)~(0.010~m)^2$
$F = 3.08~N$
(c) $F = P_g~A$
$F = [\rho~g~(h-0.20~m)](\pi~r^2)$
$F = [(800~kg/m^3)~(9.80~m/s^2)~(1.20~m-0.20~m)](\pi)~(0.0125~m)^2$
$F = 3.85~N$
(d) $F = P_g~A$
$F = [\rho~g~(h-0.20~m)](\pi~r^2)$
$F = [(800~kg/m^3)~(9.80~m/s^2)~(1.00~m-0.20~m)](\pi)~(0.0125~m)^2$
$F = 3.08~N$
(e) $F = P_g~A$
$F = [\rho~g~(h-0.20~m)](\pi~r^2)$
$F = [(1000~kg/m^3)~(9.80~m/s^2)~(1.45~m-0.20~m)](\pi)~(0.010~m)^2$
$F = 3.85~N$
We can rank the barrels in order of the force on the plug due to the liquid in the barrel, from largest to smallest:
$c = e \gt b = d \gt a$
