1. Given information

   • $m$ is the arithmetic mean of $l$ and $n$, so $m=\frac{l+n}{2}.$
   • $G_1,G_2,G_3$ are three geometric means between $l$ and $n$.

   Therefore, $l,G_1,G_2,G_3,n$ are in a geometric progression.

2. Write the GP terms

   Let the common ratio be $r$. Then $G_1=lr,\quad G_2=lr^2,\quad G_3=lr^3,\quad n=lr^4.$

   Hence, $r^4=\frac{n}{l}.$

3. Compute the required expression

   We need $G_1^4+2G_2^4+G_3^4.$

   Using the GP expressions: $G_1^4=(lr)^4=l^4r^4,$ $G_2^4=(lr^2)^4=l^4r^8,$ $G_3^4=(lr^3)^4=l^4r^{12}.$

   So, $G_1^4+2G_2^4+G_3^4=l^4r^4+2l^4r^8+l^4r^{12}.$

   Factor: $=l^4r^4(1+2r^4+r^8)=l^4r^4(1+r^4)^2.$

4. Substitute $r^4=\frac{n}{l}$

   $l^4r^4(1+r^4)^2=l^4\cdot \frac{n}{l}\left(1+\frac{n}{l}\right)^2.$

   Simplify: $=l^3n\left(\frac{l+n}{l}\right)^2$ $=l^3n\cdot \frac{(l+n)^2}{l^2}$ $=ln(l+n)^2.$

5. Use the arithmetic mean relation

   Since $m=\frac{l+n}{2},$ we have $l+n=2m.$

   Therefore, $ln(l+n)^2=ln(2m)^2=4lm^2n.$

6. Match with options

   $G_1^4+2G_2^4+G_3^4=4lm^2n.$

   This is Option D.

Final Answer: $\boxed{4lm^2n}$