Answer
$\color{red}{\bf Ethyl\; alcohol}$.
Work Step by Step
To identify a liquid, we need to find its density.
As we see in the figure below, the net force exerted on the sphere is zero since it is in a static equilibrium since it is neutrally buoyant.
Thus,
$$\sum F_y=F_B-m_sg=m_sa_y=m(0)=0$$
$$F_B=m_sg\tag 1$$
According to Archimedes’ principle, we know that the magnitude of the buoyant force equals the weight of the fluid displaced by the object.
So,
$$F_B=m_Lg_L $$
where $m_s$ is the sphere's mass and $m_L$ is the liquid's mass.
According to the density law, we know that $m=\rho V$, so
$$F_B=\rho_LV_Lg$$
Plugging into (1) and solving for $\rho_L$;
$$\rho_LV_L\color{red}{\bf\not} g=m_s\color{red}{\bf\not} g$$
Hence,
$$\rho_L=\dfrac{m_s}{V_L}$$
And since the sphere is fully submerged in the liquid, the volume of the displaced liquid is equal to the sphere's volume; $V_s=V_L$ and the volume of the sphere is $4\pi R^3/3$
$$\rho_L=\dfrac{m_s}{V_s}=\dfrac{m_s}{\frac{4}{3}\pi R_s^3}$$
Plugging the known;
$$\rho_L =\dfrac{(89.3\times 10^{-3})}{\frac{4}{3}\pi (3\times 10^{-2})^3}=789.5\;\rm kg/m^3$$
$$\rho_L \approx \bf 7.9\times 10^2\;\rm kg/m^3$$
The liquid density is about 790 kg/m$^3$ which is, as seen in table 15.1, the $\color{red}{\bf Ethyl\; alcohol}$.
