1. Form the arithmetic progression (A.P.)

If $m$ arithmetic means are inserted between $3$ and $243$, then the full A.P. has:

• first term $a=3$
• last term $l=243$
• total number of intervals $=m+1$

So the common difference is $d=\frac{243-3}{m+1}=\frac{240}{m+1}.$

The $4$th A.M. means the $4$th term inserted between $3$ and $243$. So it is the term: $3+4d=3+4\cdot \frac{240}{m+1}=3+\frac{960}{m+1}.$

2. Form the geometric progression (G.P.)

If three geometric means are inserted between $3$ and $243$, then the full G.P. is: $3,\ G_1,\ G_2,\ G_3,\ 243$

Thus there are $5$ terms total, so $243=3r^4$ where $r$ is the common ratio.

Hence $r^4=\frac{243}{3}=81=3^4$ so $r=3$ (taking the positive ratio since all terms are positive).

Therefore the G.P. is $3,\ 9,\ 27,\ 81,\ 243.$ So the $2$nd G.M. is $27.$

3. Use the given condition

Given: $\text{4th A.M.} = \text{2nd G.M.}$

So, $3+\frac{960}{m+1}=27.$

Subtract $3$: $\frac{960}{m+1}=24.$

Thus, $m+1=\frac{960}{24}=40.$ So, $m=39.$

4. Final answer

$\boxed{39}$

5. Comparison with stored answer

Stored correct answer = $39$.

Our derived answer also is $39$, so they agree.