What volume of a 3.00 m ki stock solution would you use to make 0.195 l of a 1.25 m ki solution?

For the purpose we will use solution dilution equation:
c1xV1=c2xV2
Where, c1 - concentration of stock solution; V1 - a volume of stock solution needed to make the new solution; c2 - final concentration of new solution; V2 - final volume of new solution.
c1 = 3.00 M
c2 = 1.25 M
V1 = ?
V2 = 0.195 L
When we plug values into the equation, we get following:
3 x V1 = 1.25 x 0.195
V1 = 0.08 L

About 8 gallons of fuel will be used. (8.08)