9514 1404 393
Answer:
3×10^8 m/s
Step-by-step explanation:
The desired speed is ...
[tex]\dfrac{\dfrac{9.45\times10^{15}\text{ m}}{\text{yr}}}{\dfrac{3.15\times10^7\text{ s}}{\text{yr}}}=\dfrac{9.45}{3.15}\times10^{15-7}\text{ m/s}=\boxed{3.00\times10^8\text{ m/s}}[/tex]
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