We need to solve $1+10\operatorname{Re}\left(\frac{2\cos\theta+i\sin\theta}{\cos\theta-3i\sin\theta}\right)=0,$ for $\theta\in[0,2\pi]$.

Let $z=\frac{2\cos\theta+i\sin\theta}{\cos\theta-3i\sin\theta}.$ To find $\operatorname{Re}(z)$, rationalize the denominator: $z=\frac{(2\cos\theta+i\sin\theta)(\cos\theta+3i\sin\theta)}{\cos^2\theta+9\sin^2\theta}.$

Expand the numerator: \begin{align*} (2\cos\theta+i\sin\theta)(\cos\theta+3i\sin\theta) &=2\cos^2\theta+6i\cos\theta\sin\theta+i\cos\theta\sin\theta+3i^2\sin^2\theta \ &=2\cos^2\theta-3\sin^2\theta+7i\cos\theta\sin\theta. \end{align*} So, $\operatorname{Re}(z)=\frac{2\cos^2\theta-3\sin^2\theta}{\cos^2\theta+9\sin^2\theta}.$

Now use the given equation: $1+10\cdot \frac{2\cos^2\theta-3\sin^2\theta}{\cos^2\theta+9\sin^2\theta}=0.$ Multiply through by $\cos^2\theta+9\sin^2\theta$: $\cos^2\theta+9\sin^2\theta+10(2\cos^2\theta-3\sin^2\theta)=0.$ Simplify: $\cos^2\theta+9\sin^2\theta+20\cos^2\theta-30\sin^2\theta=0,$ $21\cos^2\theta-21\sin^2\theta=0,$ $\cos^2\theta=\sin^2\theta.$ Hence, $\tan^2\theta=1.$ So, $\theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}$ in $[0,2\pi]$.

Therefore, $A=\left\{\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\right\}.$ Now compute $\sum_{\theta\in A}\theta^2=\left(\frac{\pi}{4}\right)^2+\left(\frac{3\pi}{4}\right)^2+\left(\frac{5\pi}{4}\right)^2+\left(\frac{7\pi}{4}\right)^2.$ Factor out $\frac{\pi^2}{16}$: $=\frac{\pi^2}{16}(1^2+3^2+5^2+7^2)=\frac{\pi^2}{16}(1+9+25+49)=\frac{84\pi^2}{16}.$ Thus, $\sum_{\theta\in A}\theta^2=\frac{21}{4}\pi^2.$

Comparing with the options, this is Option A.