We are given:

• $\alpha$ = number of heads in 2 tosses of coin $C_1$, where $P(H)=\frac{2}{3}$.
• $\beta$ = number of heads in 2 tosses of coin $C_2$, where $P(H)=\frac{1}{3}$.

The quadratic polynomial is $x^2-\alpha x+\beta.$ For its roots to be real and equal, the discriminant must be zero: $\alpha^2-4\beta=0.$ So we need $\alpha^2=4\beta.$

Now find possible values of $\alpha$ and $\beta$. Since each is the number of heads in 2 tosses, $\alpha,\beta\in\{0,1,2\}.$

Check which pairs $(\alpha,\beta)$ satisfy $\alpha^2=4\beta.$ Test possible values of $\alpha$:

• If $\alpha=0$, then $\alpha^2=0$, so $4\beta=0 \Rightarrow \beta=0$.
• If $\alpha=1$, then $\alpha^2=1$, so $4\beta=1$, impossible.
• If $\alpha=2$, then $\alpha^2=4$, so $4\beta=4 \Rightarrow \beta=1$.

Hence favorable cases are: $(\alpha,\beta)=(0,0) \quad \text{or} \quad (2,1).$

Compute probabilities for $\alpha$. Since $\alpha\sim \text{Binomial}(2,\tfrac23)$:

• $P(\alpha=0)=\left(\frac13\right)^2=\frac19.$
• $P(\alpha=2)=\left(\frac23\right)^2=\frac49.$

Compute probabilities for $\beta$. Since $\beta\sim \text{Binomial}(2,\tfrac13)$:

• $P(\beta=0)=\left(\frac23\right)^2=\frac49.$
• $P(\beta=1)=2\cdot \frac13\cdot \frac23=\frac49.$

Since tosses of $C_1$ and $C_2$ are independent, $\alpha$ and $\beta$ are independent. Therefore, $P(\text{real and equal roots})=P(\alpha=0,\beta=0)+P(\alpha=2,\beta=1).$

So, $P= P(\alpha=0)P(\beta=0)+P(\alpha=2)P(\beta=1).$

Substitute values: $P=\frac19\cdot\frac49+\frac49\cdot\frac49$ $=\frac{4}{81}+\frac{16}{81}$ $=\frac{20}{81}.$

Therefore the correct option is $\boxed{\frac{20}{81}}$ which is option B.