Correct Answer: $\frac{m^{2}}{3}$

Solution : Given: $a+b+c=m$
Squaring both sides, we get,
⇒ $(a+b+c)^{2}=m^{2}$
⇒ $a^{2}+b^{2}+c^{2}+2(ab+bc+ca)=m^{2}$ ------(1) [Using $(a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2(ab+bc+ca)$ ]
$\because\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$
⇒ $\frac{bc+ac+ab}{abc}=0$
⇒ $bc+ac+ab=0$
Putting the value of $bc+ac+ab=0$ in equation (1), we have,
⇒ $a^{2}+b^{2}+c^{2}+2(0)=m^{2}$
⇒ $a^{2}+b^{2}+c^{2}=m^{2}$
Dividing by 3, both sides to find the average,
⇒ $\frac{a^{2}+b^{2}+c^{2}}{3}=\frac{m^{2}}{3}$
Hence, the correct answer is $\frac{m^{2}}{3}$.