A pipe

From

$\displaystyle J_p = 2 \pi \int^{r}_{0} x^3 \ dx = \frac{\pi r^4}{2}$ ,

and

$\tau_{max} = M_k \frac{r}{J_p} = \frac{M_k}{W_p} \le k_{k}$ ,

We have

$M_k = W_k \times k_k$ ,

$\displaystyle J_p = \frac{\pi}{32} (d^{4}_{2} - d^{4}_{1}) = \frac{\pi \cdot d^{4}_{2}}{32} (1 - (\frac{d_1}{d_2})^4)$ .

$\displaystyle W_k = \frac{J_p}{\frac{d_2}{2}} = \frac{ \frac{\pi \cdot d^{4}_{2}}{32} (1- {0.8}^4)}{\frac{d_2}{2}} = \frac{ \pi}{16} \cdot d^{3}_{2} \cdot (1 - 0.4096)$ .

Plugging in, we get

$\displaystyle 500 = \frac{ \pi}{16} \cdot d^{3}_{2} \cdot 0.5904 \cdot 400$ .

$\displaystyle \implies \color{#D61F06}{d_2}$ $= \sqrt[3]{\frac{500 \cdot 16}{\pi \cdot 400 \cdot 0.6}} \approx \sqrt[3]{11} = 2.2 \ \text{cm}$ .

$\displaystyle \implies \color{#3D99F6}{d_1}$ $= 0.8 \times 2.2 = 1.76 \ \text {cm}$ .

$\displaystyle \implies$ $\color{#3D99F6}d_1$ $+ \ \color{#D61F06}d_2$ $= 1.76 + 2.2 = \boxed{3.96}$