Answer
$$P=695 \mathrm{\ kips}$$
Work Step by Step
Allowable strain in each material:
Steel:$$
\varepsilon_{s} =\frac{\sigma_{s}}{E_{s}}=\frac{20 \times 10^{3} \mathrm{psi}}{29 \times 10^{6} \mathrm{psi}}=6.8966 \times 10^{-4} $$
Concrete:
$$ \varepsilon_{c} =\frac{\sigma_{c}}{E_{c}}=\frac{2.4 \times 10^{3} \mathrm{psi}}{4.2 \times 10^{6} \mathrm{psi}}=5.7143 \times 10^{-4} $$
Smaller value governs.
$$ \varepsilon =\frac{\delta}{L}=5.7143 \times 10^{-4} $$
Let
$\quad \quad P_c$ = Portion of load stand by concrete.
$\quad \quad P_s$ = Portion of load stand by 6 steel rods.
$$\delta=\frac{P_{c} L}{E_{c} A_{c}} \therefore \quad P_{c}=E_{c} A_{c}\left(\frac{\delta}{L}\right)=E_{c} A_{c} \epsilon$$
$$\delta=\frac{P_{s} L}{E_{s} A_{s}} \therefore \quad P_{s}=E_{s} A_{s}\left(\frac{\delta}{L}\right)=E_{s} A_{s} \epsilon $$
$$ A_{s}=6\left(\frac{\pi}{4}\right) d_{s}^{2}=\frac{6 \pi}{4} (1.125 \mathrm{in} .)^{2}=5.9641 \mathrm{in}^{2} $$
$$A_{c}=\left(\frac{\pi}{4}\right) d_{c}^{2}-A_{s}=\frac{\pi}{4}(18 \mathrm{in.})^{2}-5.9641 \mathrm{in}^{2}=2.4851 \times 10^{2} \mathrm{in}^{2} $$
$$P=P_{c}+P_{s}=E_{c} A_{c} \epsilon+E_{s} A_{s} \epsilon$$
$$P=\left[\left(4.2 \times 10^{6} \mathrm{psi}\right)\left(2.4851 \times 10^{2} \mathrm{in}^{2}\right)+\\\left(29 \times 10^{6} \mathrm{psi}\right)\left(5.9641 \mathrm{in}^{2}\right)\right]\left(5.7143 \times 10^{-4}\right)$$
$$P=6.9526 \times 10^{5} \mathrm{Jb}$$
$$\therefore P=695 \mathrm{\ kips}$$
