Video solution:

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Written solution:

Let $r$ be the number of lattice points on the side length of $R.$ Similarly define $s$ for $S$ and $t$ for $T.$ Note that the number of lattice points in a rectangle is the product of the number of lattice points along its width and the number of lattice points along its length.

The first conditions gives us that $$r^2 = \dfrac{9}{4} \cdot s^2$$$$r = \dfrac{3}{2} \cdot s \tag*{(1)}$$

The number of lattice points in $R \cup T$ is the sum of the lattice points in each of the regions, but there is overlap along the $y$-axis where $S$ touches it.

The second condition, therefore, yields $$t^2 = \dfrac{1}{4}(r^2 + s^2 - s)$$ $$t^2 = \dfrac{1}{4}(\dfrac{9}{4} \cdot s^2 + s^2 - s)$$ $$t^2 = \dfrac{1}{4} \cdot \dfrac{13s^2 - 4s}{4}$$ $$16t^2 = s(13s - 4).$$ From $(1),$ we get that $s$ is a multiple of $2.$ We can substitute $s$ with $2j$ to get $$16t^2 = 2j(26j - 4)$$ $$4t^2 = j(13j - 2).$$ For the product to be divisible by $4,$ $j$ must be divisible by $2.$ We can again substitute $j$ with $2k$ to get $$4t^2 = 2k(26k - 2)$$ $$t^2 = k(13k - 1) \tag*{(2)}$$

Let $x$ be the number of lattice points along the bottom of the rectangle formed by $S \cap T$ and $y$ be the number of lattice points along the bottom of the rectangle formed by $R \cap T.$

Using these variable, we get that the number of lattice points in $S \cap T$ is $xt$ and in $R \cap T$ is $yt.$

The third condition gives us that $$\dfrac{xt}{s^2} = 27 \cdot \dfrac{yt}{r^2}$$ $$\dfrac{x}{s^2} = 27 \cdot \dfrac{y}{\dfrac{9}{4} s^2} x = 12y.$$

We also know that $t = x + y - 1$ (accounting for overlap), and this yields $$t = 13y - 1 \tag*{(3)}$$

$(3)$ gives us that $$t \equiv -1 \bmod{13} t^2 \equiv 1 \bmod{13}.$$

However, by $(2),$ we get that $$t^2 \equiv k \cdot -1 \bmod{13}$$ $$k \equiv -1 \bmod{13}.$$

By $(2),$ we also get that $k$ is a perfect square since it is relatively prime to $13k - 1,$ and they must multiply to a perfect square.

Using these restrictions on $k,$ we can try to find the smallest $k$ that works. We get that $k = 25$ satisfies both conditions.

From this value of $k,$ we get that $j = 2 \cdot 25 = 50,$ $s = 2 \cdot 50 = 100,$ and $r = \dfrac{3}{2} \cdot 100 = 150.$ We can also find that $$t^2 = 25(13 \cdot 25 - 1) = 25 \cdot 324$$ $$t = 5 \cdot 18 = 90.$$ Therefore, $$r + s + t = 340.$$ The question, however, asked for the sum of the side lengths. The side lengths of the squares are $1$ less than the number of lattice points on the side, so we have to subtract $3.$

Therefore, the desired answer is $340 - 3 = 337.$

Thus, B is the correct answer.