Let $P$ be the probability of getting an even number. Then by hypothesis the probability of getting an odd number is $3 P$. Since the events of getting an even number and an odd number are mutually exclusive and exhaustive.
$$
\therefore \quad P+3 P=1 \Rightarrow P=1 / 4
$$
Thus, the probability of getting an odd number in a single throw is $\frac{3}{4}$ and that of an even number is $\frac{1}{4}$.
If the die is thrown twice, then the sum of the numbers in two throws is even if both the numbers are even or both are odd.
$$
\therefore \text { Required probability }=\frac{3}{4} \times \frac{3}{4}+\frac{1}{4} \times \frac{1}{4}=\frac{10}{16}=\frac{5}{8}
$$