#### Answer

The density of the liquid is $890~kg/m^3$

#### Work Step by Step

We can find the volume of liquid that is displaced by the tube. Let $h$ be the length of the part of the tube that is submerged.
$V = \pi~r^2~h$
$V = (\pi)(0.020~m)^2~(0.25~m)$
$V = 3.14\times 10^{-4}~m^3$
The buoyant force on the tube is equal to the total weight of the tube and pellets. We can find the density of the liquid.
$F_B = M_{tube}~g+M_{pellets}~g$
$\rho~V~g = M_{tube}~g+M_{pellets}~g$
$\rho = \frac{M_{tube}+M_{pellets}}{V}$
$\rho = \frac{(0.030~kg)+(0.25~kg)}{3.14\times 10^{-4}~m^3}$
$\rho = 890~kg/m^3$
The density of the liquid is thus $890~kg/m^3$.