Answer
$1.8$
Since the hydrostatic pressure at the bottom of the barrel is doubled due to the presence of the narrow tube, the corresponding hydrostatic force is also doubled. But the gravitational force on the water contained in the barrel remains unchanged. Thus the the hydrostatic force on the bottom of the barrel is not equal to the gravitational force on the water contained in the barrel and the ratio is not equal to $1.0$.
Work Step by Step
The hydrostatic pressure at the bottom of the barrel is given by
$P=\rho_{w}g(L+H)$
The hydrostatic force on the bottom of the barrel is given by
$F=PA_b$
or, $F=\rho_{w}g(L+H)\pi r_b^2$
The gravitational force on the water contained in the barrel is
$W=mg$
or, $W=\rho_wVg$
or, $W=\rho_wHA_bg$
or, $W=\rho_wH\pi r_b^2g$
Therefore, the ratio yields
$\frac{F}{W}=\frac{\rho_{w}g(L+H)\pi r_b^2}{\rho_wH\pi r_b^2g}$
or, $\frac{F}{W}=\frac{(1.8+1.8)}{1.8}$
or, $\frac{F}{W}=1.8$
Therefore, the ratio of the hydrostatic force on the bottom of the barrel to the gravitational
force on the water contained in the barrel is $1.8$
Since the hydrostatic pressure at the bottom of the barrel is doubled due to the presence of the narrow tube, the corresponding hydrostatic force is also doubled. But the gravitational force on the water contained in the barrel remains unchanged. Thus the the hydrostatic force on the bottom of the barrel is not equal to the gravitational force on the water contained in the barrel and the ratio is not equal to $1.0$.
