Answer
(a) $T = 1.5\times 10^5~N$
(b) $T = 1.8\times 10^5~N$
Work Step by Step
(a) Let $M$ be the mass of the hull. Let $\rho_s$ be the density of steel. We can find the volume $V$ of the hull.
$V = \frac{M}{\rho_s}$
$V = \frac{18,000~kg}{7.8\times 10^3~kg/m^3}$
$V = 2.31~m^3$
We can find the tension $T$ in the cable. The tension will be equal to the apparent weight of the hull, which is the actual weight minus the buoyancy force. Let $\rho_w$ be the density of water.
$T = Mg - F_B$
$T = Mg - V~\rho_w~g$
$T = (18,000~kg)(9.80~m/s^2) - (2.31~m^3)(1.00\times 10^3~kg/m^3)(9.80~m/s^2)$
$T = 1.5\times 10^5~N$
(b) We can find the tension $T$ in the cable when the hull is completely out of the water.
$T = Mg$
$T = (18,000~kg)(9.80~m/s^2)$
$T = 1.8\times 10^5~N$