A uniform string of fixed length is stretched horizontally between two rigid supports and is held under tension by a block of density 1800 kg m^-3…

The block hangs in equilibrium, so the tension equals its apparent weight. In air $T_1=mg$. Fully immersed, buoyancy $\rho_w V g$ acts upward, so $T_2=mg\left(1-\dfrac{\rho_w}{\rho}\right)=mg\left(1-\dfrac{1000}{1800}\right)=\dfrac49\,mg$. For a string of fixed length $L$ and linear density $\mu$ vibrating with $p$ loops, $f=\dfrac{p}{2L}\sqrt{T/\mu}$. For the same frequency, $p\sqrt{T}$ is constant, hence $p_2=p_1\sqrt{T_1/T_2}=2\sqrt{9/4}=3$. So the string resonates with $N=3$ loops. Note that only the density ratio matters, not the mass or the volume of the block. Therefore the loop number rises because the reduced tension lowers the wave speed $v=\sqrt{T/\mu}$, so the wavelength $\lambda=v/f$ shrinks and more half-wavelengths $L=p\lambda/2$ fit into the same length. The ratio of wave speeds is $2/3$, which is exactly the ratio $p_1/p_2$, confirming $N=3$ (an integer, as it must be for a fixed-fixed string).