Answer
$ 10.2 \, \mathrm{m/s^2} $
Work Step by Step
With positive $y$-direction up, Newton's second law provides the relationship:
$$ \Sigma F_y = T - mg = ma \Rightarrow a = \frac{T}{m} - g $$
Where $T$ is the tension in the cable. We know the allowed stress for the cable, so we can use that $ \mathrm{Stress} = T/A$ to find the tension $ T = \mathrm{Stress} \times A$ and plug that in:
$$ a = \frac{\mathrm{Stress} \times A}{m} - g = $$
$$ \frac{(0.800 \times 10^8 \, \mathrm{Pa})(3.00 \times 10^{-4} \, \mathrm{m^2})}{1200 \, \mathrm{kg}} - 9.80 \, \mathrm{m/s^2} = 10.2 \, \mathrm{m/s^2}$$
