$(1)$
\[
T=\frac{D}{2}(P_1-P_2)
\]
$(2)$
\[
M=l\sqrt{(P_1+P_2)^2+(mg)^2}
\]
$(3)$
\[
Z_p=\frac{\pi}{16}d^3
\]
$(4)$
\[
\tau=\frac{T_e}{Z_p}
\]
$(5)$
\[
\tau=\frac{1}{Z_p}\sqrt{M^2+T^2}\\
=\frac{16}{\pi d^3}\sqrt{M^2+T^2}
\]
したがって,
\[\begin{align}
d&\geqq\sqrt[3]{\frac{16}{\pi\tau_a}\sqrt{M^2+T^2}}\\
&=\scriptsize{\sqrt[3]{\frac{16}{\pi\times100\times10^6}\sqrt{1.2^2\left\{(12\times10^3+8\times10^3)^2+(300\times9.8)^2\right\}+\frac{1}{4}(12\times10^3-8\times10^3)^2}}}\\
&=0.107\ \rm{m}
\end{align}
\]