In the question I know how to calculate marginal distribution for $X_1$ and $X_2$ Like for $X_1$ the marginal distribution for each column is the sum of each joint probability mass function in that column. For example the marginal distribution for $0$ column of $X_1 = 0.343$ But I'm not able to understand how to use marginal Probability of $X_1$ in the binomial distribution as asked in the question in the image!
For $X_1$~$Bi(3,p)$ the probability for $X_1=3$ is $\binom{3}{0}p^3(1-p)^0 = p^3$ therefore $p^3=0.027$ which will give $p=0.3$
$X_1 \sim Bi(3,0.3)$ is the answer & similarly we can find $X_2 \sim Bi(3,0.1)$.