\because a +b+c=0

Using the square of a trinomial formula

(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ac)

\therefore a^2+b^2+c^2+2(ab+bc+ac) = 0

\because a^2+b^2+c^2=1

\therefore ab+bc+ac = -\dfrac{1}{2}

Square both sides of the equation

a^2b^2+b^2c^2+a^2c^2+2abc(a+b+c) = \dfrac{1}{4}

Therefore,

a^2b^2+b^2c^2+a^2c^2 = \dfrac{1}{4}

Then,

a^4+b^4+c^4 = ( a^2+b^2+c^2)^2 - 2( a^2b^2+b^2c^2+a^2c^2)

= 1 - 2\times \dfrac{1}{4} = \dfrac{1}{2}

C is the choice