$\textbf{B}$

Let $r,s$ and $t$ be the number of lattice points on the side length of square $R,S$ and $T$, respectively. Note that the lengths of the side lengths are the number of lattice points minus $1$, so the sum of side lengths the question asked is $r+s+t-3$. Furthermore, the number of lattice points inside a rectangular region is equal to the number of lattice points in its width times the number of lattice points along its length. Using this fact, the number of lattice points in $R$ is $r^2$, the number of lattice points in $S$ is $s^2$, and the number of lattice points in $T$ is $t^2$.

Now, by the first condition that $R$ contains $\dfrac{9}{4}$ as many lattice points as does $S$, we have\[r^2=\frac{9}{4}\cdot s^2 \implies r = \frac{3}{2}s \quad \quad \quad \quad \quad (1)\]
The second condition, the number of lattice points contained in $T$ is $\dfrac14$ of the number of lattice points contained in $R \cup S$. The number of lattice points in $R \cup S$ is equal to the sum of the lattice points in their individually bounded regions, but the lattice points along the $y$-axis for the full length of square $S$ is shared by both of them, so we need to subtract that out. In all, this condition yields $$t^2 = \frac{1}{4}\cdot(r^2 + s^2 - s )\implies t^2 = \frac{1}{4}\cdot\left(\frac{9}{4}\cdot s^2 + s^2 - s \right)\implies 16t^2= s(13s-4)$$ Note from $(1)$ that $s$ is a multiple of $2$. We can write $s=2j$ and substitute: $$16t^2=2j(26j-4) \implies 4t^2=j(13j-2)$$ Note that $j$ must be divisible by two for the product to be divisible by 4. Thus we make another substitution, $j=2k$:\[4t^2=2k(26k-2) \implies t^2 = k(13k-1) \quad \quad \quad \quad \quad (2)\]
Then we look at the last condition that the fraction of the lattice points inside $S$ that are inside $S \cap T$ is $27$ times the fraction of lattice points inside $R$ that are inside $R \cap T$. 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 the rectangle formed by $R \cap T$. Therefore, the number of lattice points in $S\cap T$ is $xt$ and the number of lattice points in $R \cap T$ is $yt$. Thus, by this condition, $$\frac{xt}{s^2} = 27 \cdot \frac{yt}{r^2} \implies \frac{x}{s^2} = 27 \cdot \frac{y}{\frac{9}{4}\cdot s^2} \implies x= 12y$$ Notice that $t=x+y-1$, so we have\[t=x+y-1=12y+y-1=13y-1 \quad \quad \quad \quad \quad (3)\]
Now notice that by $(3)$ , $$t\equiv -1 \pmod{13}\implies t^2 \equiv 1 \pmod{13}$$
However, by $(2)$ , $t^2 \equiv -k \pmod{13}$. Therefore, $$-k \equiv 1 \pmod{13} \implies k \equiv -1 \pmod{13}$$
By Euclids algorithm, the greatest common divisor of $a$ and $b$ is $\gcd(a,b)=\gcd(b, a\mod b)$. Hence, $k$ is relatively prime to $13k-1$ because $$\gcd(13k-1,k)=\gcd(k, 13k-1\mod k)=\gcd(k,-1)=1$$
Therefore, by $(2)$ , we know $k$ must be a perfect square since $k$ is relatively prime to $13k-1$ and the two must multiply to a perfect square. Hence, we know two conditions on $k$, and we can now guess and check to find the smallest that satisfies both.

We check $k=12$ first since its one less than a multiple of $13$, but this does not work. Next, we have $k=25$ which works because $25$ is a perfect square. Thus, we have found the smallest $k$, and therefore the smallest $r, s, t$.

Now we just work backwards: $j= 2k = 50$ and $s=2j=100$. Then $r=\dfrac{3}{2}\times 100 = 150$. By $(2)$ , $t^2=25(13\times25-1) \implies t=90$.

Finally, the sum of the side lengths of each squares is $r+s+t-3=150+100+90-3=337$.