I was reading Functional Analysis- Spectral theory by V S Sunder. This particular exercise asks what you can say about the norm of the multiplicative identity if it exists.
So from the definition of a normed algebra, if we denoted multiplicative identity by $1$ and some non-zero element x $1x=x\implies ||x||\leq||1||||x||\implies 1\leq ||1||$ .
I guess that the norm of multiplicative identity will be $1$. But I am not able to derive the other inequality.
If someone can help, it will be highly appreciated.
You're not able to derive the other inequality because you can't. In general, it isn't true that the multiplicative identity has norm $1$. Let's observe, first of all, that if $V$ is a normed space and if $p: V \to \mathbb{R}$ is a norm on $V$, then $\alpha p$ is a norm on $V$ with $\alpha > 0$ being a constant.
In particular, just consider $\mathbb{R}$ with the function $p(x) = \alpha |x|$, for any $x \in \mathbb{R}$ and for some constant $\alpha > 1$. Since this is an equivalent norm, $\mathbb{R}$ is clearly a Banach space under this norm. Moreover: $$p(xy) = \alpha |xy| = \alpha |x| \cdot |y| \leq \alpha |x| \cdot \alpha |y| = p(x) p(y)$$ This shows that $\mathbb{R}$ is a Banach algebra with its usual multiplication. But now, let $x = 1$. Then, $p(1) = \alpha > 1$. So, $\mathbb{R}$ has an identity but the norm of this identity is not $1$. Of course, it is possible to take an arbitrary Banach algebra with identity and talk about its unitization; there's an equivalent norm that you can place on it which turns it into a unital Banach algebra. This construction should be available in most monographs on functional analysis that go through this stuff.