(a)
Without temperature change:
$\sigma = \dfrac{P}{A} = \dfrac{1200}{0.25} = 4800 \, \text{ psi }$
$\sigma = 4.8 \, \text{ksi} \lt 10 \, \text{ksi}$
A drop of temperature is needed to increase the stress to 10 ksi. See figure below.
$\delta = \delta_T + \delta_{st}$
$\dfrac{\sigma \, L}{E} = \alpha \, L \, (\Delta T) + \dfrac{PL}{AE}$
$\sigma = \alpha \, E \, (\Delta T) + \dfrac{P}{A}$
$10\,000 = (6.5 \times 10^{-6})(29 \times 10^6)(\Delta T) + \dfrac{1200}{0.25}$
$\Delta T = 27.59^\circ \text{F}$
Required temperature: (temperature must drop from 40°F)
$T = 40 - 27.59 = 12.41^\circ \text{F}$ answer
(b) Temperature at which the stress will be zero:
From the figure below:
$\delta = \delta_T$
$\dfrac{PL}{AE} = \alpha \, L \, (\Delta T)$
$P = \alpha AE(T_f \, - \, T_i)$
$1200 = (6.5 \times 10^{-6})(0.25) (29 \times 10^6)(T_f - 40)$
$T_f = 65.46^\circ\text{F}$ answer
