1. Let the G.P. have first term $a$ and common ratio $r$.

   Then the terms are: $a,\, ar,\, ar^2,\, ar^3,\, ar^4,\, ar^5,\, ar^6,\, ar^7,\dots$

2. Given:

   • Sum of 2nd, 3rd, and 4th terms is $3$: $ar+ar^2+ar^3=3$ $ar(1+r+r^2)=3 \qquad (1)$

   • Sum of 6th, 7th, and 8th terms is $243$: $ar^5+ar^6+ar^7=243$ $ar^5(1+r+r^2)=243 \qquad (2)$

3. Divide equation $(2)$ by equation $(1)$: $\frac{ar^5(1+r+r^2)}{ar(1+r+r^2)}=\frac{243}{3}$ $r^4=81=3^4$

   Since it is a positive term G.P., we must have $r>0$. Hence, $r=3$

4. Substitute $r=3$ into $(1)$: $a\cdot 3(1+3+9)=3$ $3a\cdot 13=3$ $39a=3$ $a=\frac{1}{13}$

5. Now find the sum of the first $50$ terms: $S_{50}=a\frac{r^{50}-1}{r-1}$

   Using $a=\frac{1}{13}$ and $r=3$, $S_{50}=\frac{1}{13}\cdot \frac{3^{50}-1}{3-1}$ $S_{50}=\frac{1}{13}\cdot \frac{3^{50}-1}{2}$ $S_{50}=\frac{1}{26}(3^{50}-1)$

6. Compare with the options: $\boxed{\frac{1}{26}(3^{50}-1)}$ This is Option D.

Therefore, the sum of the first $50$ terms is: $\boxed{\frac{1}{26}(3^{50}-1)}$