First we note that, since we have the small gray cone, which is part of the larger cone formed when appending the frustrum, we have that $$ \frac{h+h_1}{r_2} = \frac{h}{r} = 5$$ Next, we relate the volumes as follows: \begin{align} V_{\text{little cone}} &= V_{\text{frustrum}} \\\ &= V_{\text{big cone}} - V_{\text{little cone}} \\\ 2V_{\text{little cone}} &= V_{\text{big cone}} \\\ 2\cdot \frac{1}{3}\pi r^2 h &= \frac{1}{3}\pi r_2^2(h+h_1) \\\ 2r^2 h &= r_2^2(h+h_1) \\\ 2r^2 (5r) &= r_2^2(5r_2) \\\ 2r^3 &= r_2^3 \\\ r_2 &= \sqrt[3]{2} r \\\ &= \sqrt[3]{2} (100 \text{ m}) \\\ &\approx 125.992 \text{ m} \end{align} From the first equation, we have \begin{align} h_1 &= 5r_2 - h \\\ &= 5(100\sqrt[3]{2} \text{ m}) - 500 \text{ m} \\\ &= 500(\sqrt[3]{2}-1) \text{ m} \\\ &\approx 129.960 \text{ m} \end{align}