Let the missing scores be $a$ and $b$. The non-missing scores have mean $20$, so the missing scores must add up to $40$.
The standard deviation, if we use "$n-1$", not $n$, is equal to
$$\sqrt{\frac{1}{9} (226+(a-20)^2+(b-20)^2)}.$$
This is equal to $6$, so some manipulation gives
$$324=226+(a-20)^2+(b-20)^2.$$
Let $x=a-20$ and $y=b-20$. We get $x+y=0$ and $x^2+y^2=98$. Substitute. We get $x^2=49$. So $x=7$ and $y=-7$, or the other way around. Our missing values are $13$ and $27$.
Expected Value
In general, the expected value is determined by this following expression
$$\mathrm{E}(X) = \int_{-\infty}^{\infty} xf(x)\,dx$$
where $f(x)$ is the probability density function. For your problem, the expected value is
$$\begin{aligned}
\mathrm{E}(X) &= \int_0^1 x\cdot 3x^2\,dx\\
&= \int_0^1 3x^3\,dx\\
&= \left.\dfrac{3}{4}x^4\right\vert_{x = 0}^{x = 1}\\
&= \dfrac{3}{4}
\end{aligned}$$
Variance
Recall that the variance is
$$\mathrm{Var}(X) = \mathrm{E}(X^2) - (\mathrm{E}(X))^2$$
We already know $\mathrm{E}(X)$. We then need to compute $\mathrm{E}(X^2)$, which is
$$\begin{aligned}
\mathrm{E}(X^2) &= \int_0^1 x^2\cdot 3x^2\,dx\\
&= \int_0^1 3x^4\,dx\\
&= \left.\dfrac{3}{5}x^5\right\vert_{x = 0}^1\\
&= \dfrac{3}{5}
\end{aligned}$$
So
$$\mathrm{Var}(X) = \dfrac{3}{5} - \left(\dfrac{3}{4} \right)^2 = \dfrac{3}{80}$$
Standard Deviation
Thus, the standard deviation is
$$\sigma = \sqrt{\mathrm{Var}(X)} = \sqrt{\dfrac{3}{80}}$$
Best Answer
Hints:
$\mathbb E(U+V)=\mathbb E(U)+\mathbb E(V)$.
$\mathbb E(cU)=c\mathbb E(U)$ where $c$ denotes a constant.