Correct Answer: $\frac{3}{4}$

Solution : Given: $a^2+b^2+c^2=ab+bc+ca$
So, $2a^2+2b^2+2c^2-2a b-2b c-2c = 0$
⇒ $(a-b)^2+(b-c)^2+(c-a)^2=0$
If some of the positive terms are 0, the terms are themselves 0.
⇒ $(a-b)=(b-c)=(c-a)=0$
⇒ $a=b=c$
So, $\frac{11a^4+13b^4+15c^4}{16a^2b^2+19b^2c^2+17c^2a^2}$
= $\frac{11a^4+13a^4+15a^4}{16a^2a^2+19a^2a^2+17a^2a^2}$
= $\frac{11+13+15 }{16+19+17}$
= $\frac{39}{52}$
= $\frac{3}{4}$
Hence, the correct answer is $\frac{3}{4}$.