Answer
0.71 N, 984.2 g.
Work Step by Step
Three vertical forces act on the ball: the string tension, the buoyant force, and its weight.
The net force is 0.
$$F_{net}=F_T+F_b-mg=0$$
$$F_T=mg-F_b=\frac{4 \pi r^3\rho_{Cu}}{3}-\frac{4 \pi r^3\rho_{water}}{3}\approx\;0.71\;N$$
The water pushes up on the ball with a buoyant force. By Newton’s third law, the downward force on the water due to the ball is equal in magnitude to the buoyant force.
Calculate the mass equivalent of that force, $m_b$, to find the increase in the balance reading.
$$ F_b=\frac{4 \pi r^3\rho_{water}}{3} $$
$$m_b=\frac{F_b}{g}=9.203\times10^{-3}kg=9.203\;grams$$
The new reading is the old reading, 975.0 grams, plus this, or 984.2 grams.
