The correct answer is $\boxed{{1 \over 2}}$.
The convolution of two sequences $h[n]$ and $g[n]$ is defined as
$$y[n] = \sum_{k=-\infty}^{\infty} h[k] g[n-k]$$
where $u[n]$ is the unit step function.
In this case, $h[n] = {\left( {\frac{1}{2}} \right)^n} u[n]$ is a causal sequence, which means that $h[n] = 0$ for $n < 0$.
The convolution of a causal sequence with an arbitrary sequence is also causal. This means that $y[n] = 0$ for $n < 0$.
We are given that $y[0] = 1$ and $y[1] = {\frac{1}{2}}$. Using the definition of convolution, we can write
$$1 = \sum_{k=-\infty}^{\infty} h[k] g[-k]$$
and
$${\frac{1}{2}} = \sum_{k=-\infty}^{\infty} h[k] g[-k-1]$$
Subtracting these two equations, we get
$${\frac{1}{2}} = \sum_{k=-\infty}^{\infty} h[k] (g[-k-1] – g[-k])$$
Since $h[k] = 0$ for $k < 0$, the only terms that contribute to the sum are $k = 0$ and $k = -1$. Therefore, we have
$${\frac{1}{2}} = h[0] g[-1] – h[-1] g[0]$$
Solving for $g[1]$, we get
$$g[1] = \frac{h[0] + h[-1]}{{2}} = \frac{1 + {\frac{1}{2}}}{{2}} = {1 \over 2}$$