Answer
$12.2\;\rm cm$
Work Step by Step
Since the frequency of the fifth harmonic with the sphere submerged exactly matches the frequency of the third harmonic before the sphere was submerged,
$$f_{3i}=f_{5f}$$
where $_i$ is for the initial state when the sphere was in the air and $_f$ is for the final state when the sphere was submerged.
Recalling that $$f_m=\dfrac{mv}{2L}=\dfrac{m}{2L}\sqrt{\dfrac{T_s}{\mu}}$$
Plugging into (1)
$$\dfrac{3}{ \color{red}{\bf\not} 2 \color{red}{\bf\not} L}\sqrt{\dfrac{T_{si}}{ \color{red}{\bf\not} \mu}}=\dfrac{5}{ \color{red}{\bf\not} 2 \color{red}{\bf\not} L}\sqrt{\dfrac{T_{sf}}{ \color{red}{\bf\not} \mu}}$$
Squaring both sides;
$$9T_{si}=25T_{sf}\tag 1$$
And since the pulley is frictionless, the tension force in the string equals the weight of the sphere.
$$T_{si}=Mg\tag 2$$
And when the sphere is submerged,
$$T_{sf}=Mg-F_B$$
where $F_B$ is the buoyant force exerted on the sphere upward by the water, and it is given by Archimedes' principle. So, $F_B=\rho Vg$ where $\rho$ is the density of the water, $V$ is the volume of the displaced water which is the same volume of the sphere since it is fully submerged. Hence,
$$T_{sf}=Mg-\rho_{water} Vg\tag 3$$
Plugging (2) and (3) into (1);
$$9M \color{red}{\bf\not} g=25(M \color{red}{\bf\not} g-\rho_{water} V \color{red}{\bf\not} g)$$
$$9M=25M-25\rho_{water} V$$
where $V=\frac{4}{3}\pi R^3$ where $R$ is the radius of the sphere.
$$9M=25M-\frac{100\pi}{3}\rho_{water} R^3$$
Solving for $R$;
$$R=\sqrt[3]{\dfrac{25M-9M}{\frac{100\pi}{3}\rho_{water}}}$$
$$R=\sqrt[3]{\dfrac{16M}{\frac{100\pi}{3}\rho_{water}}}$$
Plugging the known;
$$R=\sqrt[3]{\dfrac{16(1.5)}{\frac{100\pi}{3}(1000)}}=\bf0.0612 \;\rm m$$
Hence, the diameter of the sphere is then
$$D=2R=2\cdot 0.0613=\bf 0.1224\;\rm m$$
$$D=\color{red}{\bf12.2}\;\rm cm$$
