The inflation rate for the period from year $k$ to year $k+1$ is the $\frac{C_{k+1}}{C_k}-1$, where in general $C_i$ is the cost at time $i$ of a "representative package" of goods. The package is meant to reflect the spending of a "representative" entity, often a family. There is a related but not identical notion of instantaneous inflation rate.
The details are quite complicated, and inflation rates in widely separated years are difficult to compare, since the "representative package" changes over time. Increases in the cost of living are only roughly captured by the inflation rate.
But let us put the above caveats aside, and compute. We are not given enough information. We are told that the inflation rate was $1.9\%$ at a certain time, and that it is $1.7\%$ now. That says nothing about the inflation rate in the period 2002-2003: it could have been $200\%$.
However, let's average the two given percentages, and assume that the inflation rate in the period from 1994 to 2014 was a constant year to year $1.8\%$. Then a representative package of goods has its price multiplied by $1.018$ each year. Thus such a package, if it cost $14000$ in 1994, should cost about $(14000)(1.018)^{20}$ today.
The calculator gives $20002.47$. This kind of precision is absurd. We can say it is about $20000$.
Remarks: $1.$ Please note that we said "representative package of goods," not "car." Things get more complicated when we deal with individual goods, for the nature of the goods, and of other goods, changes over time. Most of us who are not so young recall buying a computer around 1994, probably for more than $\$2000$. A new device with much more power costs under $\$250$ today.
$2.$ In general, suppose that we have constant inflation rate $r$ over a period of $n$ years. If for example the inflation rate is $12\%$, then $r=0.12$. Then if a representative package of goods initially costs $A$ dollars, then after $n$ years it should cost
$$A(1+r)^n.$$
