I have the following question
I have designed the following
A Binary String is even if it is ending with 0 and odd if its ending with 1.I have applied this.Im i right ?
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Your answer is incorrect. It rejects strings that are even and therefore should be accepted, like $1100$, and accepts strings that are odd and therefore should be rejected, like $1$.
The correct automaton needs to remember whether the last character read was $1$ or $0$, so we need two states: the initial state and an accepting state. When a $0$ is read, transfer to the accepting state or remain there if we were in the accepting state already. When a $1$ is read, transfer to the initial state or remain there if we were in the initial state already.
When the entire string has been read, if the last character was $0$, we'll be in the accepting state and we'll accept the string as even. If the last character was $1$, we'll be in the initial state and reject the string as odd.
UPDATE: apart from the missing initial state annotation, it looks correct after your update. Well done!