You have:
$19.6 = A e^{-2k}$
and
$19.02 = A e^{-5k}$
The above is a system of two equations in two unknowns. Surely, you can take it from here to find $A$ and $k$.
If you are not allowed the use of calculators then you can use the following strategy:
Divide the two equations to get:
$e^{3k} = \frac{19.6}{19.02}$
Now, $\frac{19.6}{19.02} \approx 1.03$ and therefore $k$ has to be less than $\frac{1}{3}$ as $e > 2$. Therefore, we can approximate $e^{3k}$ using Taylor series upto two decimals by:
$e^{3k} \approx 1 + 3k + \frac{(3k)^2}{2}$
Thus, we have:
$1 + 3k + \frac{(3k)^2}{2} = 1.03$
Solve the quadratic to identify $k$ (be careful to rule out any spurious solutions for $k$). You can then get to $A$ using the same approximation.
In fact, solving the above quadratic for $k$ yields $0.009854$ and $-0.676$ as the two possible solutions. Note that $k$ cannot be less than $0$ for the original equation as $e^{3k}$ is increasing in $k$ and $e^0 < \frac{19.6}{19.02}$. Therefore, $k=0.009854$. The accurate solution for $k$ is given by $\frac{1}{3} \mathrm{log}(\frac{19.6}{19.02})=0.0100$.