Answer
$$9.1 \times 10^{5} \mathrm{m} / \mathrm{s}$$
Work Step by Step
We know that the Equation $19-35$ gives most probable speed from where the molar mass $M:$
$$v_{p}=\sqrt{2 R T / M} . \quad $$ $$\text{With} \quad T=1 \times 10^{8} \mathrm{K} \quad $$ $$\text{and} \quad M=2.0 \times 10^{-3} \mathrm{kg} / \mathrm{mol}$$, This yields
$$v_{p}=\sqrt{\frac{2 R T}{M}}=\sqrt{\frac{2(8.314 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K})(108 \mathrm{K})}{2.0 \times 10^{-3} \mathrm{kg}}}=9.1 \times 10^{5} \mathrm{m} / \mathrm{s}$$
