1. Write the G.P. terms in standard form

Let the G.P. be $a_1=a,\, a_2=ar,\, a_3=ar^2,\, a_4=ar^3,\, a_5=ar^4$ with $a>0$ and $r>1$ since the terms are increasing positive numbers.

2. Use the condition $a_3a_5=729$

We have $a_3a_5=(ar^2)(ar^4)=a^2r^6=(ar^3)^2=729$ So, $ar^3=27$

3. Use the condition $a_2+a_4=\frac{111}{4}$

$ar+ar^3=a(r+r^3)=\frac{111}{4}$ Factor: $ar(1+r^2)=\frac{111}{4}$ Since $ar^3=27$, we get $ar=\frac{27}{r^2}$ So $\frac{27}{r^2}(1+r^2)=\frac{111}{4}$ $27\left(\frac{1+r^2}{r^2}\right)=\frac{111}{4}$ $27\left(1+\frac{1}{r^2}\right)=\frac{111}{4}$ Multiply by $4$: $108\left(1+\frac{1}{r^2}\right)=111$ $108+\frac{108}{r^2}=111$ $\frac{108}{r^2}=3$ $r^2=36$ Since $r>1$, $r=6$

4. Find $a$

From $ar^3=27$ we get $a\cdot 6^3=27$ $216a=27$ $a=\frac{1}{8}$

Thus, $a_1=\frac18,\quad a_2=\frac68=\frac34,\quad a_3=\frac{36}{8}=\frac92$

5. Compute $24(a_1+a_2+a_3)$

$a_1+a_2+a_3=\frac18+\frac34+\frac92$ Convert to denominator $8$: $=\frac18+\frac68+\frac{36}{8}=\frac{43}{8}$ Therefore, $24(a_1+a_2+a_3)=24\cdot \frac{43}{8}=3\cdot 43=129$

6. Check options

The value is $\boxed{129}$ So the correct option is B.