Correct Answer: 32

Solution : Given, $8 a^3+27 b^3=16$ and $2 a+3 b=4$
Consider $8 a^3+27 b^3=16$
⇒ $(2a)^3+(3b)^3=16$
⇒ $(2a+3b)(4a^2-6ab+9b^2)=16$
⇒ $(4)(4a^2-6ab+9b^2)=16$
⇒ $4a^2-6ab+9b^2=4$ ......(1)
Now, consider $2a+3b=4$
Squaring both sides, we get,
⇒ $(2a+3b)^2=4^2$
⇒ $4a^2+12ab+9b^2=16$ ......(2)
Solving (1) and (2), we get,
$18 ab=12$
⇒ $ab=\frac{2}{3}$
And, $4a^2+9b^2=4+6\times\frac23$
⇒ $4a^2+9b^2=4+4=8$
Now, $16 a^4+81 b^4$
$=(4a^2)^2+(9b^2)^2$
$=(4a^2+9b^2)^2-72a^2b^2$
$=8^2-72\times\frac{2}{3}\times \frac{2}{3}$
$=64-32=32$
Hence, the correct answer is 32.