Answer
$0.11N$
Work Step by Step
We can find the required weight as follows:
$P=\frac{F}{A}=\frac{643}{\pi(0.75)^2}=1455Pa$
As $P=\rho gh$
$\implies h=\frac{P}{\rho g}=\frac{1455}{(1000)(9.81)}=0.148$
Now $W=mg=\rho(\pi (\frac{d}{2})^2)hg$
We plug in the known values to obtain:
$W=1000(\pi(\frac{0.010}{2})^2)(0.148)(9.81)$
$W=0.11N$
