It's a bit easier to solve this problem by flipping the curve on its side; i.e., instead of looking at $y=-x^2+25$:
$\hskip2.7in $ 
looking at $y=\sqrt{25-x}$:
$\hskip1.4in $ 
Using the latter setup, what you want to find first (if I have understood your question correctly) are the non-negative values of $x$ where $$y=\sqrt{25-x}=\frac{n}{3}$$
where $n$ is an integer (all my numbers will be in feet, hence $4$ inches is $\frac{1}{3}$). These are just the numbers
$$25-\frac{n^2}{9}$$
as $n$ ranges from $1$ to $15$, i.e.
224/9, 221/9, 24, 209/9, 200/9, 21, 176/9, 161/9, 16, 125/9, 104/9, 9, 56/9, 29/9, 0
or, to use approximate values,
24.89, 24.56, 24.00, 23.22, 22.22, 21.00, 19.56, 17.89, 16.00, 13.89, 11.56, 9.000, 6.222, 3.222, 0
This shows how the number of wood pieces you can fit within the boundary increases by one at each of these values of $x$:

As you can see, for $x\leq25-\frac{n^2}{9}$, we can fit in $n$ wood pieces.
Now, the thickness of each wood piece is 2 inches, or $\frac{1}{6}$, so the top of the $m$th row (counting the row whose bottom is the $y$-axis as row 1) is at $x=\frac{m}{6}$. Thus, you will be able to fit $n$ wood pieces in on the $m$th row if and only if
$$\frac{m}{6}\leq25-\frac{n^2}{9},$$
or
$$n\leq \sqrt{225-\frac{3m}{2}},$$
so the number of wood pieces you can fit in on the $m$th row is
$$\left\lfloor\sqrt{225-\frac{3m}{2}}\right\rfloor.$$
Using this formula, we can generate a side-on view of what the final result of your project will look like:

Is this what you had in mind?
Mathematica code for that final image:
f[m_] := Floor[Sqrt[225 - (3 m/2)]]
Show[ParametricPlot[Table[{m/6, u*f[m]/3}, {m, 1, 150}], {u, 0, 1}],
Plot[Sqrt[25 - x], {x, 0, 25}, AspectRatio -> 1/5],
ParametricPlot[Table[{u Floor[150 - 2 n^2/3]/6, n/3}, {n, 1, 15}], {u, 0, 1}],
PlotRange -> {0, 5}]