The individual will invest the whole $\\\$ 10,000$, which makes sense, since in the worst case scenario he still gets a positive return. He has to decide which fraction $\alpha$ to invest in asset $A$ and which fraction $1-\alpha$ to invest in asset $B$.
Note that if returns are the same for both assets, the fraction invested on each is irrelevant. Then, his problem can be written as follows,
$$
\begin{aligned}
\max _{\alpha} B= & \frac{1}{10} \sqrt{1.2 \times 10,000}+\frac{1}{10} \sqrt{1.05 \times 10,000} \\
& +\frac{4}{10} \sqrt{1.2 \times \alpha 10,000+1.05 \times(1-\alpha) 10,000} \\
& +\frac{4}{10} \sqrt{1.05 \times \alpha 10,000+1.2 \times(1-\alpha) 10,000}
\end{aligned}
$$
Let's simplify the problem first.
Since there are two situations in which $\alpha$ does not affect the outcome, we can rewrite the problem as follows,
$$
\begin{aligned}
\max _{\alpha} \tilde{B}= & \frac{4}{10} \sqrt{1.2 \times \alpha 10,000+1.05 \times(1-\alpha) 10,000} \\
& +\frac{4}{10} \sqrt{1.05 \times \alpha 10,000+1.2 \times(1-\alpha) 10,000} \\
\max _{\alpha} B= & \frac{4 \sqrt{10,000}}{10}[\sqrt{1.2 \alpha+1.05(1-\alpha)}+\sqrt{1.05 \alpha+1.2(1-\alpha)}]
\end{aligned}
$$
Then, we can just solve,
$$
\max _{\alpha} B=\sqrt{1.05+0.15 \alpha}+\sqrt{1.2-0.15 \alpha}
$$
The first order condition is given by:
$$
\frac{\partial B}{\partial \alpha}=0.15(1.05+0.15 \alpha)^{-\frac{1}{2}}-0.15(1.2-0.15 \alpha)^{-\frac{1}{2}}=0
$$
Then,
$$
\begin{aligned}
(1.05+0.15 \alpha)^{-\frac{1}{2}} & =(1.2-0.15 \alpha)^{-\frac{1}{2}} \\
1.05+0.15 \alpha & =1.2-0.15 \alpha \\
0.3 \alpha & =0.15
\end{aligned}
$$
Then, the $\alpha=0.5$.
This is not surprising, given that the assets are symmetric. Let us finally verify that the second order condition holds:
$$
\frac{\partial^{2} B}{\partial \alpha^{2}}=-\frac{1}{2} 0.15^{2}(1.05+0.15 \alpha)^{-\frac{3}{2}}-\frac{1}{2}(-0.15)^{2}(1.2-0.15 \alpha)^{-\frac{3}{2}}<0
$$
