Let $$z_0$$ be a root of quadratic equation, $$x^2 + x + 1 = 0$$. If $$z = 3 + 6iz_0^{81} - 3iz_0^{93}$$, then $$\arg(z)$$ is equal to:

We start with the quadratic equation $$x^{2}+x+1=0$$. Its root $$z_{0}$$ therefore satisfies

$$z_{0}^{2}+z_{0}+1=0.$$

Now, multiply both sides by $$z_{0}-1$$ (this is a standard trick for such quadratics):

$$\left(z_{0}^{2}+z_{0}+1\right)(z_{0}-1)=0.$$

Expanding the left‐hand side term by term:

$$z_{0}^{3}-z_{0}^{2}+z_{0}^{2}-z_{0}+z_{0}-1=0.$$

Observe that $$-z_{0}^{2}+z_{0}^{2}=0$$ and $$-z_{0}+z_{0}=0$$, so we are left with

$$z_{0}^{3}-1=0.$$

Hence

$$z_{0}^{3}=1.$$

Because the quadratic had no real roots (its discriminant $$\Delta=1-4=-3<0$$), we know $$z_{0}\neq 1$$. Thus $$z_{0}$$ is one of the two non-real cube roots of unity, and the key property we need is the period-3 relation

$$z_{0}^{k+3}=z_{0}^{k}\quad\text{for any integer }k.$$

Now we examine the given expression

$$z=3+6i\,z_{0}^{81}-3i\,z_{0}^{93}.$$

Using the property $$z_{0}^{3}=1$$, we reduce the large exponents modulo 3. First, note the division algorithm:

$$81=3\times27+0 \quad\Longrightarrow\quad 81\equiv0\pmod{3},$$ $$93=3\times31+0 \quad\Longrightarrow\quad 93\equiv0\pmod{3}.$$

So

$$z_{0}^{81}=z_{0}^{0}=1, \qquad z_{0}^{93}=z_{0}^{0}=1.$$

Substituting these results back into the expression for $$z$$ we obtain

$$z = 3 + 6i\,(1) - 3i\,(1) = 3 + 6i - 3i.$$

Simplifying the imaginary part, we arrive at

$$z = 3 + 3i.$$

To find the argument, we recall the definition: for a complex number $$a+ib$$ in the first quadrant,

$$\arg(a+ib)=\tan^{-1}\!\left(\dfrac{b}{a}\right).$$

Here, $$a=3$$ and $$b=3$$, so

$$\arg(z)=\tan^{-1}\!\left(\dfrac{3}{3}\right)=\tan^{-1}(1)=\dfrac{\pi}{4}.$$

Hence, the correct answer is Option B.