Determine g(t) for minimal symbol error probability, given h(t), K=1 and Minimum distance receiver
Your computation using the energy of $h(t)$, $E_h$, to find the value of $c$, is correct. But note that you are only forcing the zero ISI condition for $k=0$. You still need to check the zero ISI criterion whenever $k\neq0$, but there is no mention of the value of $T$, and we can't derive what happens in those cases until we know the value of $T$.
For a general approach, I'd recommend going to the basics and not taking shortcuts. One key problem in your definition is that your convolution integral is not taking into account the time instant $t$, and therefore, you only compute one value of the convolution (for $t=0$), instead of all the values that are relevant (namely, $kT$ for all $k$).

The convolution of $h(t)$ and $g(t)$ is defined as
$$p(t) = (g* h)(t) = \int_{-\infty}^{\infty}g(\tau)h(t-\tau)\,d\tau = \int_{-\infty}^{\infty}h(\tau)g(t-\tau)\,d\tau$$
and we need to satisfy the Nyquist first criterion for zero ISI,
$$p(t)\Big|_{t=kT}=(g* h)(kT)=\left\{\begin{array}{rl}1,&amp;k=0\\0,&amp;k\neq0\end{array}\right.$$
Lastly, the matched filter is defined as $g(t)=ch^*(T-t)$ in its causal version, and $g(t)=ch^*(-t)$ in its non-causal version. For simplicity, let's use the noncausal version:
$$\begin{align}
p(t)&amp; = (g* h)(t) = c\int_{-\infty}^{\infty}h(\tau)h^*(\tau-t)\,d\tau\\
\end{align}$$
In the particular case of $t=0$ (equivalently, $k=0$), we have
$$\begin{align}
p(0)&amp; = (g* h)(0) = 1\\
&amp;=c\int_{-\infty}^{\infty}h(\tau)h^*(\tau)\,d\tau= c\int_{-\infty}^{\infty}|h(\tau)|^2\,d\tau\\
&amp;= c\Bigg(\int_{0}^{10^{-3}}|2|^2\,d\tau+\int_{10^{-3}}^{2\cdot10^{-3}}|1|^2\,d\tau\Bigg)=c\big(4\cdot10^{-3}+10^{-3}\big)=5c\cdot10^{-3}\\
&amp;\Rightarrow c=10^3/5 = 200
\end{align}$$
Now, for verifying that the system satisfies the zero ISI criterion, we need the value of $T$. But you can follow a similar procedure.