Correct Answer: $24\sqrt 3$ cm

Solution :
Given: AL, BM, and CN are altitude of $\triangle$ ABC.
AL, BM, and CN are medians of $\triangle$ ABC.
Since ABC is an equilateral triangle,
AO = 2OL
⇒ 16 = 2OL
⇒ OL = 8
$\therefore$ AL = AO + OL = 16 + 8 = 24 cm
Let AB = BC = CA = $x$ cm
In $\triangle$ ALC we have,
$\angle$ ALC = 90$^\circ$ ($\because$ AL is altitude)
AL = 24 cm
AC = $x$ cm
CL = $\frac{\text{BC}}{2}$ = $\frac{x}{2}$
$\therefore x^2$ = $(\frac{x}{2})^2+24^2$
⇒ $x^2$ = $\frac{x^2}{4}+576$
⇒ $\frac{3x^2}{4}$ = $576$
⇒ $x^2$ = $\frac{576×4}{3}$
⇒ $x^2$ = $768$
⇒ $x$ = $16\sqrt3$
Semi perimeter = $\frac{16\sqrt3+16\sqrt3+16\sqrt3}{2}$
= $\frac{48\sqrt3}{2}$
= $24\sqrt3$
Hence, the correct answer is $24\sqrt3$ cm.