You assumed that $C$ was the initial amount ("concentration" is a term indicating the ratio of "active" amount to total amount, which is not something so easily controlled). But this is not the case. The initial amount is $N(0)$, and that value is $$N(0) = C + \frac a\lambda - \frac b{\lambda^2}$$
When $a = 300, b = 20, C = 300$ and the half-life is $200$ years, that value is about
$-1578233.5$. The reason that after 10 years you have a negative amount is that you started with an even more negative amount and have not yet managed to overcome it.
If you want an initial amount $N(0)$ to be $300$, then you need to set $C = 300 - \frac a\lambda + \frac b{\lambda^2} \approx 1578833.5$
However, as I indicated in a comment, I am doubtful this equation is modelling what you think it is.
By your description, the repository starts out with an amount of $300$. The first year, an additional $300$ are added to it, for a total of $600$, not including decay. The next year, an additional $320$ are added, the year after $340$ are added, then $360$ the year after that, etc. Since you described it as "linear", I don't think this is what you are after.
Instead, I am guessing that $a$ is supposed to be the initial amount, and each year an additional amount $b$ is added. This gives the differential equation
$$\frac {dN}{dt} = b -\lambda N$$ whose solution is
$$N(t) = \frac b{\lambda} + Ce^{-\lambda t}$$
Since $a = N(0) = b/\lambda + C$, we have $C = a - b/\lambda$. Therefore
$$N(t) = \frac b{\lambda} + \left(a - \frac b\lambda\right)e^{-\lambda t}$$
Is the equation you are after.
Plug in $t=0$ into the solution you get, you will have $x=Ae^0=A$. So you have value for $A$. For $k$, as you have shown, you have $0.9x=xe^{-200k}$. So you have
$$-200k=\ln(0.9).$$
When you consider the case after 1000 years, you have
$$-1000k=5\ln(0.9)=\ln(0.9^5).$$
So you will see
$$e^{-1000k}=0.9^5.$$
This is the percent you are looking for.
Best Answer
Hint: If you start with 1 unit of the isotope at time $0$, you have $\frac 12$ after $150$ years. Do you know how to solve $m'=km$? That should give you an equation that (with this data) lets you determine $k$. Now put that $k$ into your solution and find the time when the amount is $0.15$.
Added: your equation is the correct differential equation. The solution is $m(t)=m(0)\exp(-kt)$ as the derivative of $\exp(-kt)$ is $-k \exp(-kt)$