If $\mathrm{m}$ arithmetic means (A.Ms) and three geometric means (G.Ms) are inserted between $3$ and $243$ such that $4^{th}$ A.M. is equal to $2^{nd}$ G.M., then $m$ is equal to:

Let $3, A_1, A_2, A_3,......A_m, 243$ are in arithmetic progression with $m$ arithmetic means.

Common difference $d=\frac{243-3}{m+1}=\frac{240}{m+1}$

Let $3, G_1, G_2, G_3, 243$ are in geometric progression with $3$ geometric means.

Common ratio $r={\left(\frac{243}{3}\right)}^{\frac{1}{3+1}}={\left(81\right)}^{\frac{1}{4}}=3$

Given $G_2=A_4$

$\Rightarrow 3{\left(3\right)}^2=3+4\left(\frac{240}{m+1}\right)$

$\Rightarrow 27=3+\frac{960}{m+1}$

$\Rightarrow m+1=40$

$\Rightarrow m=39$