Answer
Since a density of $7.8\times 10^3~kg/m^3$ is similar to the density of iron and steel, the sample is most likely made of iron or steel.
Work Step by Step
Let $M_a$ be the apparent mass of the sample. Let $\rho_w$ be the density of water. We can find the volume $V$ of the sample.
$M_a~g = M~g-F_B$
$F_B = M~g-M_a~g$
$\rho_w~V~g = M~g-M_a~g$
$V = \frac{M-M_a}{\rho_w}$
$V = \frac{0.0635~kg-0.0554~kg}{1.00\times 10^3~kg/m^3}$
$V = 8.10\times 10^{-6}~m^3$
We can find the density $\rho_s$ of the sample as:
$\rho_s = \frac{0.0635~kg}{8.10\times 10^{-6}~m^3}$
$\rho_s = 7.84\times 10^3~kg/m^3$
Since a density of $7.8\times 10^3~kg/m^3$ is similar to the density of iron and steel, the sample is most likely made of iron or steel.
